zoj2588 Burning Bridges(无向图的桥)

题目请戳这里

题目大意：给一张无向图，现在要去掉一些边，使图仍然连通，求不能去掉的边。

题目分析：就是求无向图的桥。

tarjan算法跑一遍，和无向图割点十分类似，这里要找low[v] > dfn[u]的边(u,v)便是割边，因为v是u的孩子，但是v无法访问到u的祖先，那么断开这条边原图必不连通，因此这是桥。这题会有平行边，平行边必定不是桥。所以dfs的时候要判断一下。

详情请见代码：

 

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 10005;
const int M = 500005;
int m,n,num,ansnum,dfns;
int head[N],ans[M],low[N],dfn[N];
bool vis[N];
struct node
{
    int to,next,id;
}bridge[M<<1];
void build(int s,int e,int id)
{
    bridge[num].id = id;
    bridge[num].to = e;
    bridge[num].next = head[s];
    head[s] = num ++;
}
void dfs(int cur,int fa)
{
    vis[cur] = true;
    int chongbian = 0;
    dfn[cur] = low[cur] = dfns ++;
    for(int i = head[cur];i != -1;i = bridge[i].next)
    {
        if(fa == bridge[i].to)
            chongbian ++;
        if(vis[bridge[i].to] == false)
        {
            dfs(bridge[i].to,cur);
            low[cur] = min(low[cur],low[bridge[i].to]);
            if(low[bridge[i].to] > dfn[cur])
                ans[ansnum ++] = bridge[i].id;
        }
        else if(fa != bridge[i].to || chongbian > 1)
                low[cur] = min(low[cur],dfn[bridge[i].to]);
    }
}
void tarjan()
{
    int i;
    dfns = 1;
    memset(vis,false,sizeof(vis));
    memset(dfn,0,sizeof(dfn));
    for(i = 1;i <= n;i ++)
        if(vis[i] == false)
            dfs(i,-1);
    printf("%d\n",ansnum);
    sort(ans,ans + ansnum);
    for(i = 0;i < ansnum;i ++)
        printf(i == ansnum - 1?"%d\n":"%d ",ans[i]);
}
int main()
{
    int t,i;
    int a,b;
    scanf("%d",&t);
    while(t --)
    {
        scanf("%d%d",&n,&m);
        memset(head,-1,sizeof(head));
        num = ansnum = 0;
        for(i = 1;i <= m;i ++)
        {
            scanf("%d%d",&a,&b);
            build(a,b,i);
            build(b,a,i);
        }
        tarjan();
        if(t)
            puts("");
    }
    return 0;
}