zoj——2588 Burning Bridges

Burning Bridges

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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Ferry Kingdom is a nice little country located on N islands that are connected by M bridges. All bridges are very beautiful and are loved by everyone in the kingdom. Of course, the system of bridges is designed in such a way that one can get from any island to any other one.

But recently the great sorrow has come to the kingdom. Ferry Kingdom was conquered by the armies of the great warrior Jordan and he has decided to burn all the bridges that connected the islands. This was a very cruel decision, but the wizards of Jordan have advised him no to do so, because after that his own armies would not be able to get from one island to another. So Jordan decided to burn as many bridges as possible so that is was still possible for his armies to get from any island to any other one.

Now the poor people of Ferry Kingdom wonder what bridges will be burned. Of course, they cannot learn that, because the list of bridges to be burned is kept in great secret. However, one old man said that you can help them to find the set of bridges that certainly will not be burned.

So they came to you and asked for help. Can you do that?

Input

The input contains multiple test cases. The first line of the input is a single integer T (1 <= T <= 20) which is the number of test cases. T test cases follow, each preceded by a single blank line.

The first line of each case contains N and M - the number of islands and bridges in Ferry Kingdom respectively (2 <= N <= 10 000, 1 <= M <= 100 000). Next M lines contain two different integer numbers each and describe bridges. Note that there can be several bridges between a pair of islands.

Output

On the first line of each case print K - the number of bridges that will certainly not be burned. On the second line print K integers - the numbers of these bridges. Bridges are numbered starting from one, as they are given in the input.

Two consecutive cases should be separated by a single blank line. No blank line should be produced after the last test case.

Sample Input

2

6 7
1 2
2 3
2 4
5 4
1 3
4 5
3 6

10 16
2 6
3 7
6 5
5 9
5 4
1 2
9 8
6 4
2 10
3 8
7 9
1 4
2 4
10 5
1 6
6 10

Sample Output

2
3 7

1
4 思路：tarjan求割边的裸题但要注意有重边，我们再输入的时候要判断一下，判断他是不是重边，如果是的话，就记录一下，方便后面找割边的时候进行判断，因为如果两个边之间有重边的话，那么这两个点之间一定没有割点。其次就是要注意输出，md我只想问谁出的这道题，（不要拦我，我想去揍他、、、），，输出那么鬼畜，害的调了整整一天、、、、代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 200100
using namespace std;
int n,m,x,y,z,s,t,tim,tot;
int dfn[N],low[N],head[N],ans_edge[N];
struct Edge
{
    int from,to,next,id,flag;
}edge[N];
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
int add(int id,int x,int y)
{
    for(int i=head[x];i;i=edge[i].next)
     ;;}
    ++tot;
    edge[tot].id=id;
    edge[tot].to=y;
    edge[tot].next=head[x];
    head[x]=tot;
    ++tot;
    edge[tot].id=id;
    edge[tot].to=x;
    edge[tot].next=head[y];
    head[y]=tot;
}
void clean()
{
    tim=,tot=;s=;
    memset(dfn,,sizeof(dfn));
    memset(low,,sizeof(low));
    memset(head,,sizeof(head));
    memset(ans_edge,,sizeof(ans_edge));
}
int tarjan(int now,int pre)
{
    ;bool boo=false;
    dfn[now]=low[now]=++tim;
    for(int i=head[now];i;i=edge[i].next)
    {
        int t=edge[i].to;
        )==pre) continue;
        if(!dfn[t])
        {
            sum++;tarjan(t,i);
            low[now]=min(low[now],low[t]);
            ) ans_edge[++s]=edge[i].id;
        }
        else low[now]=min(low[now],dfn[t]);
    }
}
int main()
{
    t=read();
    while(t--)
    {
        n=read(),m=read();clean();
        ;i<=m;i++)
          x=read(),y=read(),add(i,x,y);
        tarjan(,);
        printf("%d\n",s);
        sort(ans_edge+,ans_edge++s);
        ;i<s;i++)
         printf("%d ",ans_edge[i]);
        if(s) printf("%d\n",ans_edge[s]);
        if(t) printf("\n");
    }
    ;
}