zoj2588 Burning Bridges(无向图的桥)

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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;}