HDU 5631 Rikka with Graph

Rikka with Graph

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

題意：現在個你一個圖。有n個點，有n+1個邊，問你去下一些邊仍是連通的；

題解：由於仍是連通的，所以最多去掉兩個邊，暴力即可；

#include<stdio.h>
#include<string.h>
int road[1010];
int n;
struct node {
    int a;
    int b;
}st[110];
int find(int a)
{
    if(road[a]==a) return a;
    else
    return find(road[a]);
}
void mix(int a,int b)
{
    int x;
    int y;
    x=find(a);
    y=find(b);
    if(x!=y)
    road[x]=y;
}

 void init()
 {
        for(int i=1;i<=n;i++)
        road[i]=i;  
 }

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();

        scanf("%d",&n);
        int a, b;
        for(int i=1;i<=n+1;i++)
        {
            scanf("%d%d",&a,&b);
            st[i].a =a;
            st[i].b =b;
        }   
        int ans=0,s=0;
        for(int i=1;i<=n+1;i++)//一條邊； 
        {
            init();
            s=0;
            for(int j=1;j<=n+1;j++)
            {
                if(j!=i)
                {
                    mix(st[j].a ,st[j].b );

                }

             } 
            for(int ll=1;ll<=n;ll++)
                if(road[ll]==ll)
                {
                    s++;
                }
                if(s==1)
                ans++;

        }
    //  printf("%d||| \n",ans);
        for(int i=1;i<=n;i++) //two ways
        {
            for(int j=i+1;j<=n+1;j++)
            {
                init();
                s=0;
                for(int k=1;k<=n+1;k++)
                {
                    if(k==i||k==j)
                    {
                        continue;
                    }
                    else
                    {
                            mix(st[k].a ,st[k].b );
                    }
                }
                for(int zz=1;zz<=n;zz++)
                if(road[zz]==zz)
                {
                    s++;
                }
                if(s==1)
                ans++;
            }
        }


        printf("%d\n",ans);
    }
}