DP：求出3点构成链的方案数 ，然后总方案数减去它

Y

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#pragma comment(linker, "/STACK:16777216")

using namespace std;

typedef long long int LL;

const int maxn=100100;
vector<int> g[maxn];

int n;
LL ans,doubi[maxn],son[maxn];

void dfs(int u,int fa)
{
  LL ndoubi=0,nson=0;
  son[u]=1;
  for(int i=0,sz=g[u].size();i<sz;i++)
    {
      int v=g[u][i];
      if(v==fa) continue;
      dfs(v,u);
      son[u]+=son[v];
      doubi[u]+=doubi[v]+son[v];

      ans+=son[v]*nson;
      ans+=son[v]*ndoubi;
      ans+=doubi[v]*nson;
      ans+=doubi[v];

      nson+=son[v];
      ndoubi+=doubi[v];
    }
}

int main()
{
  while(scanf("%d",&n)!=EOF)
    {
      ans=0;
      memset(doubi,0,sizeof(doubi));
      memset(son,0,sizeof(son));
      for(int i=0;i<=n+10;i++)
        g[i].clear();
      for(int i=0;i<n-1;i++)
        {
          int a,b;
          scanf("%d%d",&a,&b);
          g[a].push_back(b);
          g[b].push_back(a);
        }
      dfs(1,1);
      cout<<1LL*n*(n-1)*(n-2)/6-ans<<endl;
    }
  return 0;
}