hdu4705 Y 2013 Multi-University Training Contest 10

Y

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 304    Accepted Submission(s): 104

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")

一定要构个无向图，不构无向图会有神数据卡你:

#pragma comment(linker, "/STACK:16777216")
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <set>
#include <vector>
#define ll long long int
using namespace std;
vector<int> a[100005];
int b[100005];
ll sum,n;
ll x[100005];
int fun(int x1)
{
    b[x1]=1;
    int size=a[x1].size();
     x[x1]=1;
    int i;
    for(i=0;i<size;i++)
    if(!b[a[x1][i]])
    x[x1]+=fun(a[x1][i]);
    return x[x1];
}
void dfs(int x1)
{
    b[x1]=1;
    int size=a[x1].size();
    if(size==0) return ;
    int i;
    ll ma=n-1;
    for(i=0;i<size;i++)
    if(!b[a[x1][i]])
    {
       ma-=x[a[x1][i]];
    }
    for(i=0;i<size;i++)
    if(!b[a[x1][i]])
    sum-=ma*x[a[x1][i]],ma+=x[a[x1][i]],dfs(a[x1][i]);
}
int main()
{
    ll x1,y1;
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        memset(x,0,sizeof(x));
        sum=n*(n-1)/2*(n-2)/3;
        memset(b,0,sizeof(b));
        for(i=1;i<=n;i++)
        a[i].clear();
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&x1,&y1);
            a[x1].push_back(y1);
            a[y1].push_back(x1);
        }
        fun(1);
        memset(b,0,sizeof(b));
        dfs(1);
        cout<<sum<<endl;
    }
}