zoj 2588 Burning Bridges 联通分量

求一个无向图的桥（可能存在重边），输出割边的数目，并按顺序输出割边的序号(输入的顺序）。

由于有重边，一般需要使用邻接表来存储，我一开始嫌麻烦，想使用邻接矩阵和边集来存，没注意到节点数目太大，结果MLE。最终还是得用邻接表，写好后，有贡献了一次PE，改正格式后居然WA了，经检查在插入边时有点问题，插入时要先查找该边是否已出现，当初使用的是if(findEdge(a,b)==0&&findEdge(b,a)==0)来判断边是否已插入，后来发现这个不能保证findEdge(a,b)和findEdge(b,a)都会执行，改成if(findEdge(a,b)+findEdge(a,b))就AC了。但至今不解当初为何会PE，OJ是按怎样的顺序判错的呢？

Description

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

26 71 22 32 45 41 34 53 610 162 63 76 55 95 41 29 86 42 103 87 91 42 410 51 66 10

Sample Output

23 714 

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;int cnt,n,s_bridge,times;int head[200010],dfn[200010],low[200010],bridge[200010];struct node{    int v,id,next;} edge[200010];void add(int u,int v,int id){    edge[cnt].v=v;    edge[cnt].next=head[u];    edge[cnt].id=id;    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].id=id;    edge[cnt].next=head[v];    head[v]=cnt++;}void tarjan(int u,int fa){    int flag=0;    dfn[u]=low[u]=++times;    for(int e=head[u]; e!=-1; e=edge[e].next)    {        int v=edge[e].v;        if(v==fa&&!flag)        {            flag=1;            continue;        }        if(!dfn[v])        {            tarjan(v,u);            low[u]=min(low[u],low[v]);            if(low[v]>dfn[u])                bridge[s_bridge++]=edge[e].id;        }        else            low[u]=min(low[u],dfn[v]);    }}int main(){    int Case;    int m;    int u,v;    scanf("%d",&Case);    while(Case--)    {        memset(head,-1,sizeof(head));        memset(dfn,0,sizeof(dfn));        times=0;        cnt=0;        scanf("%d %d",&n,&m);        for(int i=1; i<=m; i++)        {            scanf("%d %d",&u,&v);            add(u,v,i);        }        s_bridge=0;        tarjan(1,-1);        printf("%d\n",s_bridge);        if(s_bridge)        {            sort(bridge,bridge+s_bridge);            for(int i=0; i<s_bridge-1; i++)                printf("%d ",bridge[i]);            printf("%d\n",bridge[s_bridge-1]);        }        if(Case)            printf("\n");    }    return 0;}