hihocoder1479

有2种情况。都选tot/3的节点，那么，各节点记录子节点的为tot/3的乘一下。然后是祖先是tot/3*2的乘以子树中tot/3的数量。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <cassert>
#include <stack>
#include <bitset>
#include <list>
#define mkp make_pair
#define err cout<<"err"<<endl
using namespace std;
const double EPS=1e-12;
typedef long long lon;
typedef unsigned long long ull;
const lon SZ=100010,SSZ=2*SZ,APB=52,one=1;
const lon INF=0x7f7f7f7f,mod=1000000007;
int n,arr[SZ],sum[SZ],num[SZ],tot;
vector<int> mp[SZ];
lon ans;

void dfs1(int x,int p)
{
    sum[x]=arr[x];
    //cout<<"x: "<<x<<" "<<arr[x]<<endl;
    for(int i=0;i<mp[x].size();++i)
    {
        int to=mp[x][i];
        if(to!=p)
        {
            dfs1(to,x);
            sum[x]+=sum[to];
        }
    }
}

void dfs2(int x,int p)
{
    if(sum[x]==tot/3)++num[x];
    //cout<<"x: "<<x<<" "<<sum[x]<<" "<<num[x]<<endl;
    lon pre=0;
    lon tmp=(sum[x]==tot/3*2)&&(p!=-1);
    for(int i=0;i<mp[x].size();++i)
    {
        int to=mp[x][i];
        if(to!=p)
        {
            dfs2(to,x);
            num[x]+=num[to];
            ans+=pre*num[to];
            ans+=tmp*num[to];
            pre+=num[to];
        }
    }
}

void init()
{
    cin>>n;
    for(int i=1;i<=n;++i)
    {
        int a,b;
        cin>>a>>b;
        arr[i]=a;
        if(!b)continue;
        mp[i].push_back(b);
        mp[b].push_back(i);
    }
    dfs1(1,-1);
    tot=sum[1];
    if(tot%3)
    {
        cout<<0<<endl;
    }
    else
    {
        dfs2(1,-1);
        cout<<ans<<endl;
    }
}

void work()
{
    
}

void release()
{
    for(int i=1;i<=n;++i)
    {
        mp[i].clear();
        num[i]=0;
    }
    ans=0,tot=0;
}

int main()
{
    std::ios::sync_with_stdio(0);
    //freopen("d:\\1.txt","r",stdin);
    int casenum;
    cin>>casenum;
    //cout<<casenum<<endl;
    for(int time=1;time<=casenum;++time)
    //for(int tim=1;cin>>n;++tim)
    {
        init();
        work();
        release();
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/gaudar/p/10952142.html