Rikka with Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 905 Accepted Submission(s): 404

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

131 22 33 11 3

Sample Output

Source

/*思路：题目意思是一个n个点的无向图，给你n+1条边，求有多少种删除办法，使得图还能连通至少删除一条边我的思路是，n个点，要连通，至少是n-1条边，所以最多删除2条，枚举每一条边，删除求连通若可以，+1，尝试枚举下一条边被删，看能不能继续连通 */  /*  超时   没用邻接表  复杂度O（n*n）的递归  */  /*#include
    
     #include 
     
      using namespace std;const int N=100+5;struct Node{	int x,y;}node[N];int check; int map[N][N];  //记录边数   用int可以累加 bool book[N];int n;void dfs(int now){   //深搜 	book[now]=1;	check++;	for(int i=1;i<=n;i++){     //这里时间复杂度n^2 		if(book[i])          continue;		 if(map[now][i]>0)		  dfs(i);	}}int main(){      int t,re;	cin>>t;	while(t--){		  re=0;		 cin>>n;		 memset(map,0,sizeof(map));		 for(int i=0;i
      
       >node[i].x>>node[i].y;		 	 map[node[i].x][node[i].y]++;		 	 map[node[i].y][node[i].x]++;		 }		 for(int i=0;i
       
        #include 
        
          #include 
         
          #include 
          
           using namespace std;const int N=100+5;struct Node{ int x,y;}node[N];vector
           
             ve[N];int map[N][N]; //记录边数 用int可以累加 bool book[N];int check; int n;void dfs(int now){ //深搜 book[now]=1; check++; for(int i=0;i
            
             0) dfs(ve[now][i]); }}//用一个全局变量减少判断，复杂度减少O(n)//inline bool check(){ //求连通 // for(int i=1;i<=n;i++)// if(!book[i])// return false; // return true;//}int main(){ int t,re; scanf("%d",&t); while(t--){ re=0; scanf("%d",&n); memset(map,0,sizeof(map)); for(int i=1;i<=n;i++){ ve[i].clear(); } for(int i=0;i
             
              >node[i].x>>node[i].y; ve[node[i].x].push_back(node[i].y); ve[node[i].y].push_back(node[i].x); map[node[i].x][node[i].y]++; map[node[i].y][node[i].x]++; } memset(book,0,sizeof(book)); check=0; dfs(1); if(check!=n){ printf("0\n"); continue; } for(int i=0;i