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[];
 int b[];
 ll sum,n;
 ll x[];
 int fun(int x1)
 {
     b[x1]=;
     int size=a[x1].size();
      x[x1]=;
     int i;
     for(i=;i<size;i++)
     if(!b[a[x1][i]])
     x[x1]+=fun(a[x1][i]);
     return x[x1];
 }
 void dfs(int x1)
 {
     b[x1]=;
     int size=a[x1].size();
     if(size==) return ;
     int i;
     ll ma=n-;
     for(i=;i<size;i++)
     if(!b[a[x1][i]])
     {
        ma-=x[a[x1][i]];
     }
     for(i=;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,,sizeof(x));
         sum=n*(n-)/*(n-)/;
         memset(b,,sizeof(b));
         for(i=;i<=n;i++)
         a[i].clear();
         for(i=;i<n;i++)
         {
             scanf("%d%d",&x1,&y1);
             a[x1].push_back(y1);
             a[y1].push_back(x1);
         }
         fun();
         memset(b,,sizeof(b));
         dfs();
         cout<<sum<<endl;
     }
 }