HDU-4705-树形dp/组合数学

Y

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2762    Accepted Submission(s): 800

Problem Description

 

Sample Input

4 1 2 1 3 1 4

 

Sample Output

1

Hint

1. The only set is {2,3,4}. 2. Please use #pragma comment(linker, "/STACK:16777216")

 

Source

2013 Multi-University Training Contest 10

 

       可以先找到三个点的SIMPLE-PATH的数量，然后让C(N,3)-SIMPLE-PATH就是答案了。

      对于节点u，子树的节点个数为n1,n2,,,,,nk的话，以u为中点的SIMPLE-PATH的数量就是n1(n2+n3+...+nk)+n2(n3+...+nk)+.....+nk-1*nk,这一步可以在遍历儿子时迭代实现。

      别忘记还有一颗子树是父亲节点，N-sum[u].

   

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
#pragma comment(linker, "/STACK:16777216")
#define LL long long 
vector<int>g[100005];
LL sum[100005];
LL tot,N;
void dfs(int u,int pre){
    sum[u]=1;
    LL  res=0;
    for(int i=0;i<g[u].size();++i){
        if(g[u][i]==pre) continue;
        dfs(g[u][i],u);
        sum[u]+=sum[g[u][i]];
        tot+=res*sum[g[u][i]];
        res+=sum[g[u][i]];
    }
    tot+=res*(N-sum[u]);
}
int main()
{
    while(cin>>N){int u,v;
        tot=0;
        for(int i=1;i<N;++i){
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs(1,0);
        LL tt=0;
        LL ans=N*(N-1)*(N-2)/6;
        cout<<ans-tot<<endl;
        for(int i=1;i<=N;++i) g[i].clear();
    }
    return 0;
}