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Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

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(T30)——The number of the testcases. For each testcase, the first line contains a number n(n100). 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

给了n点,n+1条边,问存在多少个方案使图去掉几条边后,仍能连通。
由连通图性质知,n个点至少要有n-1条边才能连通
因为数据范围较小
所以尝试采取枚举法看去掉某一条或某两条边后图是否连通
判断图是否连通可用并查集

#include<stdio.h>
int par[120];
struct node {
    int n,m;
} a[120];
int find(int m) {
    if(m==par[m])
        return m;
    else
        return par[m]=find(par[m]);
}
void unite(int x,int y) {
    x=find(x);
    y=find(y);
    if(x==y)
        return ;
    else
        par[y]=x;
}
int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        int n;
        scanf("%d",&n);
        for(int i=0; i<n+1; i++) {
            scanf("%d %d",&a[i].n,&a[i].m);
        }
        int res=0;
        for(int i=0; i<n+1; i++) {
            for(int j=i; j<n+1; j++) {
                for(int k=1; k<=n; k++)
                    par[k]=k;
                for(int k=0; k<n+1; k++) {
                    if(k==i||k==j)
                        continue;
                    unite(a[k].n,a[k].m);
                }
                int k;
                for(k=1; k<n; k++) {
                    if(find(n)!=find(k))
                        break;
                }
                if(k==n)
                    res++;
            }
        }
        printf("%d\n",res);
    }
    return 0;
}



查看原文:<a href=http://www.boiltask.com/blog/?p=1965>http://www.boiltask.com/blog/?p=1965</a>