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;

}