Rikka with Graph||HDU5631

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：给了n个节点和n+1条边，所以若要连通可以删除一条边或者两条边，循环判定有多少满足即可

#include
#include
#include
#include
#include
#include
using namespace std;
int n,par[102],a[102],b[102],c[102];
int find(int x)//找祖宗
{
    if(x!=par[x])
      par[x]=find(par[x]);
    return par[x];
}
void unite(int x,int y)//合并 
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
      par[fx]=fy;
}
bool judge()
{
    int i,ans=0;
    for(i=1;i<=n;i++)
       par[i]=i;
    for(i=0;i<=n;i++)
    {
        if(c[i])
          unite(a[i],b[i]);
    }
    for(i=1;i<=n;i++)
    {
        if(i==par[i])
          ans++;
    }
    if(ans>1)
      return 0;
    return 1;
}
int main()
{
    int t,i,j,sum;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%d",&n);
        for(i=0;i<=n;i++)
        {
            scanf("%d%d",&a[i],&b[i]);
            c[i]=1;
        }
        for(i=0;i<=n;i++)
        {
            c[i]=0;
            if(judge())//删除一条边判断 
              sum++;
            for(j=i+1;j<=n;j++)
            {
                c[j]=0;
                if(judge())//删除两条边判断 
                  sum++;
                c[j]=1;
            }
            c[i]=1;
        }
        printf("%d\n",sum);
    } 
return 0;
}

先模仿，再独创，再超越！