UVa 11795 – Mega Man’s Mission

0

Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2895


/*
     If opportunity doesn't knock, build a door.

            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
            |S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

    Success is how high you bounce when you hit bottom.
*/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
int Set(int N,int pos){return N=N | (1<<pos);}
int reset(int N,int pos){return N= N & ~(1<<pos);}
bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

int n;
ll dp[18][(1<<17)+100];
int ara[18][18];
int temp[18];

ll func(int idx, int mask)
{
    if(__builtin_popcount(mask)==n) return 1;

    ll &ret=dp[idx][mask];

    if(ret!=-1) return ret;

    ll p=0;

    for(int i=0; i<n; i++)
    {
        if(!check(mask,i))
        {
            if(ara[idx][i] || ara[0][i] || temp[i])
            {
                vector<int>v;
                loop(j,n)
                {
                    if(temp[j]==0 && ara[idx][j])
                    {
                        temp[j]=1;
                        v.pb(j);
                    }
                }
                p+=func(i+1,Set(mask,i));
                loop(j,SZ(v)) temp[v[j]]=0;
            }
        }
    }

    return ret=p;

}



int main()
{
    CIN;
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);

    int t;
    cin>>t;
    TEST_CASE(t)
    {
        cin>>n;
        loop(i,n+1)
        {
            string str;
            cin>>str;
            loop(j,n) ara[i][j]=(str[j]=='1');
        }

        ms(dp,-1);
        CASE_PRINT;
        cout<<func(0,0)<<endl;

    }

    return 0;
}

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SPOJ ARRAYSUB – subarrays

0

Problem Link : http://www.spoj.com/problems/ARRAYSUB

Solution Idea:

This is a RMQ (Range Max/Min Query) problem. This types of problem can be solved in many ways like segment tree, sliding window etc. But in this problem time limit is very tight and for this reason we have to use Sliding window technique which complexity is O(n) for solving this problem.



/*
         +-+ +-+ +-+
         |R| |.| |S|
         +-+ +-+ +-+
 */

#include <bits/stdc++.h>

#define pii             pair <int,int>
#define sc              scanf
#define pf              printf
#define Pi              2*acos(0.0)
#define ms(a,b)         memset(a, b, sizeof(a))
#define pb(a)           push_back(a)
#define MP              make_pair
#define db              double
#define ll              long long
#define EPS             10E-10
#define ff              first
#define ss              second
#define sqr(x)          (x)*(x)
#define D(x)            cout<<#x " = "<<(x)<<endl
#define VI              vector <int>
#define DBG             pf("Hi\n")
#define MOD             100007
#define MAX             10000
#define CIN             ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)           (int)a.size()
#define sf(a)           scanf("%d",&a)
#define sfl(a)          scanf("%lld",&a)
#define sff(a,b)        scanf("%d %d",&a,&b)
#define sffl(a,b)       scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)     scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)    scanf("%lld %lld %lld",&a,&b,&c)
#define loop(i,n)       for(int i=0;i<n;i++)
#define REP(i,a,b)      for(int i=a;i<b;i++)
#define TEST_CASE(t)    for(int z=1;z<=t;z++)
#define PRINT_CASE      printf("Case %d: ",z)
#define all(a)          a.begin(),a.end()
#define intlim          2147483648
#define inf             1000000
#define ull             unsigned long long

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

int ara[1000005];
deque<int> big,small;

int main()
{
    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    CIN;
    int n,d;
    cin>>n;
    loop(i,n) cin>>ara[i];
    cin>>d;

    if(d==1)
    {
        cout<<ara[0];
        REP(i,1,n) cout<<" "<<ara[i];
        cout<<endl;
    }

    else
    {

        big.pb(0);
        int ans=0;
        bool test=0;
        REP(i,1,n)
        {
            while(!big.empty() && ara[big.back()]<=ara[i])
                big.pop_back();
            big.pb(i);
            while(i-big.front()>=d)
                big.pop_front();

            if(i>=d-1)
            {
                if(test)
                    cout<<" "<<ara[big.front()];
                else
                    cout<<ara[big.front()];
                test=1;

            }
        }

        cout<<endl;
    }
    return 0;
}


UVa 11608 – No Problem

0

Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2655


/*
     If opportunity doesn't knock, build a door.

            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
            |S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

    Success is how high you bounce when you hit bottom.
*/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<":"<<endl
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/



int main()
{
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);

    int n,s;
    int z=0;
    while(cin>>n && n>=0)
    {

        z++;
        int ara1[15],ara2[15];

        loop(i,12) cin>>ara1[i];
        loop(i,12) cin>>ara2[i];
        CASE_PRINT;
        loop(i,12)
        {
            if(n>=ara2[i])
            {
                cout<<"No problem! :D"<<endl;
                n-=ara2[i];
            }
            else
            {
                cout<<"No problem. :("<<endl;
            }
            n+=ara1[i];
        }


    }

    return 0;
}


UVa 993 – Product of digits

0

Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=934

Solution Idea:

Just divide the number by integers 9,8,7,6,5,4,3 and 2  and when we can divide the digit with a number then we can add it to a string. and at last if The last number is greater than 9 then we can’t express that number in that way. So the ans will be -1 and If we can divide the whole number by 9 to 2 then after all division we need to reverse the string because we have to print minimum number.



/*
     If opportunity doesn't knock, build a door.

            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
            |S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

    Success is how high you bounce when you hit bottom.
*/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

int main()
{
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);


    int t;
    cin>>t;
    TEST_CASE(t)
    {
        int n;
        cin>>n;

        if(n<10)
        {
            cout<<n<<endl;
            continue;
        }

        string str;
        int last=-1;
        while(n>1 && last!=n)
        {
            last=n;
            for(int i=9; i>=2; i--)
            {
                if(n%i==0)
                {
                    while(n%i==0)
                    {
                        n/=i;
                        str+='0'+i;
                    }
                }
            }
        }

        reverse(all(str));

        if(n>1)
            cout<<-1<<endl;
        else
            cout<<str<<endl;


    }

    return 0;
}

UVa 11850 – Alaska

0

Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2950

Solution Idea:

The tricky part in this problem statement is this –

” Can Brenda drive her car from Dawson City to Delta Junction and back? ”

So Brenda need to reach her destination and return his start point. So just take input all the points and sort then in acceding order and check that the distance between two consecutive points is less than or equal to 200 or not. If it’s not then the ans will be impossible and if the 2*(distance between the last point and and destination) >200 the ans will be impossible. Otherwise the ans is possible.


/*
     If opportunity doesn't knock, build a door.

            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
            |S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

    Success is how high you bounce when you hit bottom.
*/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/



int main()
{
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int n;
    while(cin>>n && n)
    {
        vector<int>v;

        loop(i,n)
        {
            int a;
            cin>>a;
            v.pb(a);
        }

        sort(all(v));

        bool test=0;

        REP(i,1,n)
        {
            if(v[i]-v[i-1]>200)
            {
                cout<<"IMPOSSIBLE"<<endl;
                test=1;
                break;
            }
        }

        if(test==1) continue;

        if((1422-v[n-1])*2>200) test=1;

        if(test==0)
            cout<<"POSSIBLE"<<endl;
        else
            cout<<"IMPOSSIBLE"<<endl;

    }
    return 0;
}

UVa 11614 – Etruscan Warriors Never Play Chess

0

Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2661

Solution Idea:

In this problem the number of warriors is given and we have to say how many row can be made with them in this rule that the row contains warriors equal to row number (i th row contains i warriors). so that if there is 3 row there we have 1+2+3 = 6 worriers.

We can use binary search to solve this problem. let low=0 and high= 10000000000 (A large value ) and binary search the ans which satisfy the formula n*(n+1)/2  according to given input.


/*
     If opportunity doesn't knock, build a door.

            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
            |S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
            +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

    Success is how high you bounce when you hit bottom.
*/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

ll ask(ll n)
{
    return n*(n+1)/2;
}

int main()
{
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int t;
    cin>>t;
    TEST_CASE(t)
    {
        ll x;
        cin>>x;
        ll lo=0,hi=10000000000,ans=0;

        while(lo<=hi)
        {
            ll mid=(lo+hi)/2;

            ll temp=ask(mid);

            if(temp<=x)
                {
                    ans=mid;
                    lo=mid+1;
                }
            else
                hi=mid-1;
        }

        cout<<ans<<endl;
    }
    return 0;
}


UVa 762 – We Ship Cheap

0

Problem Link :https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=703

Solution Idea:

After reading the problem statement It’s clear that this is a shortest path problem and we can solve this using BFS. In this problem all the nodes are string. so we use map for mapping strings with integers and run a BFS.

The main problem is is problem is it’s test cases. This problem have a lots of critical case. Like source and destination is not present on the input graph and Source and Destination is the same node.

Think carefully about this test cases and handle them separately and Most important thing in solving graph problem is clearing the whole graph and arrays after each test cases.


/*
         +-+ +-+ +-+
         |R| |.| |S|
         +-+ +-+ +-+
*/

#include <bits/stdc++.h>

#define pii             pair <int,int>
#define sc              scanf
#define pf              printf
#define Pi              2*acos(0.0)
#define ms(a,b)         memset(a, b, sizeof(a))
#define pb(a)           push_back(a)
#define MP              make_pair
#define db              double
#define ll              long long
#define EPS             10E-10
#define ff              first
#define ss              second
#define sqr(x)          (x)*(x)
#define D(x)            cout<<#x " = "<<(x)<<endl
#define VI              vector <int>
#define DBG             pf("Hi\n")
#define MOD             100007
#define MAX             10000
#define CIN             ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)           (int)a.size()
#define sf(a)           scanf("%d",&a)
#define sfl(a)          scanf("%lld",&a)
#define sff(a,b)        scanf("%d %d",&a,&b)
#define sffl(a,b)       scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)     scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)    scanf("%lld %lld %lld",&a,&b,&c)
#define loop(i,n)       for(int i=0;i<n;i++)
#define REP(i,a,b)      for(int i=a;i<b;i++)
#define TEST_CASE(t)    for(int z=1;z<=t;z++)
#define PRINT_CASE      printf("Case %d: ",z)
#define all(a)          a.begin(),a.end()
#define intlim          2147483648
#define inf             1000000
#define ull             unsigned long long

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

vector<int>G[10005];
map<string,int>mp;
map<int,string>mp1;

int d[10005];
bool visit[10005];
int dir[10005];
int cnt=0;

int bfs(int src, int dst)
{
    queue<int>Q;
    d[src]=0;
    visit[src]=1;
    Q.push(src);
    dir[src]=src;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        if(u==dst) return 1;

        loop(i,SZ(G[u]))
        {
            int v=G[u][i];
            if(!visit[v])
            {
                visit[v]=1;
                d[v]=d[u]+1;
                Q.push(v);
                dir[v]=u;
            }
        }
    }
    return -1;
}

void path_print(int a, int b)
{
   if(a==b) return;
   path_print(dir[a],b);
   cout<<mp1[dir[a]]<<" "<<mp1[a]<<endl;
}

void allclear(int n)
{
    loop(i,n)
    {
        G[i].clear();
        d[i]=visit[i]=dir[i]=0;
    }
    mp.clear();
    mp1.clear();
    cnt=0;
}

int main()
{
    CIN;
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int n;
    bool test=0;
    while(cin>>n)
    {
        if(test) cout<<endl;
        test=1;
        string s1,s2;
        loop(i,n)
        {
            cin>>s1>>s2;
            if(mp[s1]==0)
               {
                 mp[s1]= ++cnt;
                 mp1[cnt]=s1;
               }
            if(mp[s2]==0)
                {
                    mp[s2]= ++cnt;
                    mp1[cnt]=s2;
                }
            G[mp[s1]].pb(mp[s2]);
            G[mp[s2]].pb(mp[s1]);
        }

        cin>>s1>>s2;
        if(s1==s2)
            cout<<s1<<" "<<s1<<endl;
        else if((mp[s1]==0 || mp[s2]==0))
        {
            cout<<"No route"<<endl;
        }
        else if(bfs(mp[s1],mp[s2])==-1)
            cout<<"No route"<<endl;
        else
            path_print(mp[s2],mp[s1]);

    allclear(cnt+5);

    }
    return 0;
}