ZOJ 2588 Burning Bridges

求哪几条边是桥，用Tarjan算法，注意重边的处理。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

const int maxn=10000+10; //结点数量
const int Maxn=2*100000+10; //边的数量
int low[maxn];
int dfn[maxn];
struct Edge
{
    int from;
    int to;
    int id;
    int ans;//ans为1，表示这条边是割边
} edge[Maxn];
vector<int>G[maxn];
int Ans[Maxn];
int N,M;//N个结点，M条边
int tmpdfn;
int tot;

void init()
{
    for(int i=0; i<maxn; i++) G[i].clear();
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    low[1]=dfn[1]=1;
    tmpdfn=0;
    tot=0;
}

int dfs(int u,int id)
{
    tmpdfn++;
    int lowu=dfn[u]=tmpdfn;
    for(int i=0; i<G[u].size(); i++)
    {
        int B=G[u][i];
        if(!dfn[edge[B].to])
        {
            int lowv=dfs(edge[B].to,edge[B].id);
            lowu=min(lowu,lowv);
            if(lowv>dfn[u]) edge[B].ans=1;
        }
        else if(dfn[edge[B].to])
        {
            if(edge[B].id/2==id/2) continue;
            lowu=min(lowu,dfn[edge[B].to]);
        }
    }
    low[u]=lowu;
    return lowu;
}

void AddEdge(int u,int v)
{
    edge[tot].from=u;
    edge[tot].to=v;
    edge[tot].id=tot;
    edge[tot].ans=0;
    G[u].push_back(tot);
    tot++;

    edge[tot].from=v;
    edge[tot].to=u;
    edge[tot].id=tot;
    edge[tot].ans=0;
    G[v].push_back(tot);
    tot++;
}

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d%d",&N,&M);
        init();
        for(int i=0; i<M; i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            AddEdge(u,v);
        }

        dfs(1,-1);

        int ans=0;
        for(int i=0; i<2*M; i++)
            if(edge[i].ans)
            {
                Ans[ans]=edge[i].id/2+1;
                ans++;
            }
        printf("%d\n",ans);
        for(int i=0; i<ans; i++)
        {
            if(i<ans-1) printf("%d ",Ans[i]);
            else printf("%d\n",Ans[i]);
        }
        if(T_T>=1) printf("\n");
    }
    return 0;
}