UVa 12582 – Wedding of Sultan

Problem Link : http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=4027

Solution Idea:

In this problem we just need to count the number of edges connected to a node. In other words We need to calculate for all nodes that how many edges is connected to them. DFS traversal of the graph is given.  We can use a stack to calculate the ans as like DFS.


/*
         +-+ +-+ +-+
         |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 CASE_PRINT      cout<<"Case "<<z<<endl
#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[30];

int main()
{
    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int t;
    cin>>t;
    TEST_CASE(t)
    {
        stack<int>st;
        string str;

        cin>>str;
        ms(ara,0);
        st.push(str[0]-'A');
        for(int i=1; i<SZ(str); i++)
        {
            int b=str[i]-'A';
            if(st.top()==b)
            {
                st.pop();
                if(!st.empty())
                {
                    ara[st.top()]++;
                    ara[b]++;
                }
            }
            else
                st.push(b);
        }
        CASE_PRINT;
        for(int i=0; i<26; i++)
            if(ara[i])
                cout<<(char)('A'+i)<<" = "<<ara[i]<<endl;

    }
    return 0;
}

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