Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1023    Accepted Submission(s): 486

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

 

 

Source

 

 

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一开始被这道题给蒙了一下，要讨论多少种情况。。。。。

后来才知道，数据只有n+1条边，要想成树，最起码要有n-1条边，所以只有两种讨论的情况，删一条边或者删两条边。

#include
      
       #include
       
        using namespace std;int u[121], v[121], par[121];int n, ans;void init() {    for(int i = 0;i <= n;i++) {        par[i]=i;    }}int Find(int x) {    if (x == par[x])        return x;    return par[x] = Find(par[x]);}void unite(int x, int y) {    int fx = Find(x);    int fy = Find(y);    if(fx != fy) par[fx] = fy;}void dele1(int x) {    init();    for(int i=0;i
        
         <= n; i++) {        if(par[i] == i) cont++;    }    if(cont == 1) ans++;}void dele2(int x,int y) {    init();    for(int i = 0; i < n + 1; i++) {        if(i != x && i != y) {            unite(u[i], v[i]);        }    }    int cont = 0;    for(int i = 1;i <= n; i++) {        if(par[i] == i) cont++;    }    if(cont == 1) ans++;}int main() {    int t;    scanf("%d", &t);    while(t--) {        scanf("%d", &n);        for(int i = 0; i < n + 1; i++) {            scanf("%d%d", &u[i], &v[i]);        }        ans = 0;        for(int i = 0;i < n+1; i++) {            dele1(i);        }        for(int i = 0;i < n+1; i++) {            for(int j = i+1; j < n+1; j++) {                dele2(i, j);            }        }        printf("%d\n", ans);    }    return 0;}