Codeforces: 877E. Danil and a Part-time Job

Problem Link : http://codeforces.com/contest/877/problem/E

Solution Idea:

    Run a dfs on the given for getting the Euler tour order or inorder traversal order. Then the tree is converted to a liner 1D array. Now we can apply range query and update on this array.

    Now use a range update query data structure. Here I have used segment tree. Now initialize the tree with the initial value and now our job is to just toggle a subtree of the given tree and perform query operation. Use a propagation for the update and answer the query.



#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#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)             cerr<<#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); cout.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 loop1(i,n)       for(int i=1;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 LINE_PRINT_CASE  printf("Case %d:\n",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;

//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;


/*----------------------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));}
/*------------------------------------------------*/

#define mx 200005
#define segment_tree int l=n*2,r=l+1,mid=(b+e)/2
pii info[mx];
vector<int>g[mx];
int cnt=0;
int maps[mx],ara[mx];

void dfs(int u, int par)
{
    info[u].ff=++cnt;
    maps[cnt]=u;
    for(int i=0;i<SZ(g[u]);i++)
    {
        int v=g[u][i];
        if(v==par) continue;
        dfs(v,u);
    }
    info[u].ss=cnt;
}

struct data
{
    int val, prop;
};

data tree[3*mx];

void push_down(int n, int b, int e)
{
    if(tree[n].prop%2)
    {
        segment_tree;
        tree[l].val=(mid-b+1)-tree[l].val;
        tree[l].prop++;
        tree[r].val=(e-mid)-tree[r].val;
        tree[r].prop++;
        tree[n].prop=0;
    }
}

void init(int n, int b, int e)
{
    if(b==e)
    {
        tree[n].val=ara[maps[b]];
        tree[n].prop=0;
        return;
    }
    segment_tree;
    init(l,b,mid);
    init(r,mid+1,e);
    tree[n].val=tree[l].val+tree[r].val;
}

void update(int n, int b, int e, int i, int j)
{
    if(b>j || e<i) return;
    if(b>=i && e<=j)
    {
        tree[n].val=(e-b+1)-tree[n].val;
        tree[n].prop++;
        return;
    }
    push_down(n,b,e);
    segment_tree;
    update(l,b,mid,i,j);
    update(r,mid+1,e,i,j);
    tree[n].val=tree[l].val+tree[r].val;
}

int query(int n, int b, int e, int i, int j)
{
    if(b>j || e<i) return 0;
    if(b>=i && e<=j) return tree[n].val;
    push_down(n,b,e);
    segment_tree;
    int p=query(l,b,mid,i,j);
    int q=query(r,mid+1,e,i,j);
    return p+q;
}

char str[10];

int main()
{

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

    int n;
    sf(n);
    for(int i=2;i<=n;i++)
    {
        int a;
        sf(a);
        g[a].pb(i);
    }

    for(int i=1;i<=n;i++) sf(ara[i]);
    dfs(1,1);
    init(1,1,n);
    int q;
    sf(q);
    while(q--)
    {
        int a;
        scanf(" %s %d",str,&a);
        if(str[0]=='g')
        {
            int x=query(1,1,n,info[a].ff,info[a].ss);
            printf("%d\n",x);
        }
        else
        {
            update(1,1,n,info[a].ff,info[a].ss);
        }
    }



    return 0;
}

Advertisements