HDU5631 Rikka with Graph

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)T(T≤30)——The number of the testcases.

For each testcase, the first line contains a number n(n≤100)n(n≤100).

Then n+1 lines follow. Each line contains two numbers u,vu,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

Hint

题意

题解：

由于n很小 所以直接枚举1和2的情况 判断不加入某一条边或某两条边成立的个数

AC代码

#include<cstdio>#include<cstring>#include<stack>#include <set>#include <queue>#include <vector>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int mod = 1e9+7;int par[110];int u[115];int v[115];void init(int n){    for (int i = 1;i <= n; ++i){        par[i] = i;    }}int fd(int x){    if (x == par[x]) return par[x];    return par[x] = fd(par[x]);}int main(){    int t;    scanf("%d",&t);    while (t--){        int n;        scanf("%d",&n);        for (int i = 1;i <= n+1; ++i){            scanf("%d%d",&u[i],&v[i]);        }        int ans;        int num = 0;        for (int i = 1; i <= n+1; ++i){            init(n);            ans = n;            for (int j = 1;j <= n+1; ++j){                if (j!=i){                    int tx = fd(u[j]);                    int ty = fd(v[j]);                    if (tx!=ty){                        ans--;                        par[tx] = ty;                    }                }            }if (ans==1) num++;        }        ans = n;        for (int i = 1;i <= n+1; ++i){            for (int j = i+1; j <= n+1; ++j){                init(n);                ans = n;                for (int k = 1;k <= n+1; ++k){                    if (k!=i&&k!=j){                        int tx = fd(u[k]);                        int ty = fd(v[k]);                        if (tx!=ty){                            ans--;                            par[tx] = ty;                        }                    }                } if(ans==1) num++;            }        }        printf("%d\n",num);    }    return 0;}