Rikka with Graph(联通图取边，暴力)

Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 190    Accepted Submission(s): 78

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one) and delete them from the graph.
Yuta wants to know the number of the ways to choose the edges in order to make the remaining graph connected.
It is too difficult for Rikka. Can you help her?

 

Input

The first line contains a number T(T≤30)——The number of the testcases.
For each testcase, the first line contains a number n(n≤100).
Then n+1 lines follow. Each line contains two numbers u,v , which means there is an edge between u and v.

 

Output

For each testcase, print a single number.

 

Sample Input

1
3
1 2
2 3
3 1
1 3

 

Sample Output

9

 

题解：题目问的是去掉边使图仍然联通。很简单的一道图论题，我竟然用prime错了半天。。。最后还是改了krustra,让找有多少中取法，直接暴力取边

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x,y) scanf("%lf%lf",&x,&y)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=;
int pre[MAXN];
int s[MAXN],e[MAXN];
int N,ans;
int find(int r){
    return pre[r]= pre[r]==r?r:find(pre[r]);
}
int check(int a,int b){
    for(int i=;i<=N;i++)pre[i]=i;
    for(int i=;i<=N;i++){
        if(i==a||i==b)continue;
        int f1=find(s[i]),f2=find(e[i]);
        //printf("%d %d\n",f1,f2);
        if(f1!=f2)pre[f1]=f2;
    }
    int cnt=;
    for(int i=;i<=N;i++){
        if(pre[i]==i)cnt++;
    //    if(cnt>1)printf("%d\n",cnt);
        if(cnt>)return ;
    }
    return ;
}
int main(){
    int T;
    SI(T);
    while(T--){
        SI(N);
        for(int i=;i<=N;i++)
            SI(s[i]),SI(e[i]);
            int ans=;
        for(int i=;i<=N;i++)
            for(int j=i;j<=N;j++){//相等代表的是取一条边。
                ans+=check(i,j);
            }
        printf("%d\n",ans);
    }
    return ;
}