zoj 2588 Burning Bridges (割边/桥)

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题目链接: zoj 2588

题目大意: 在无向连通图中找桥，并且按序号输出

解题思路: Tarjan查找割边

数据中可能会出现重边，需要判断是否是重边

每次加入新的边，则枚举此顶点之前加入的边中是否存在另一点

low[vv]>dnf[u]是桥，low[vv]>=dnf[u]是割点

代码:

//Final  Tarjan 找桥(割边)#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>using namespace std;#define MAX 10005#define MIN(a,b) a<b?a:bstruct snode{int to,next,num,m;}edge[MAX*20];int visit[MAX],low[MAX],dnf[MAX],pre[MAX],ans[MAX],ansn,Index,n,high;void Add_edge(int a,int b,int mm){int i;for(i=pre[a];i!=-1;i=edge[i].next)  //遍历一遍之前加入该点的所有边{if(edge[i].to==b)break;}if(edge[i].to==b)    //记录重边，重边不是桥edge[i].num++;else{edge[Index].to=b;edge[Index].m=mm;edge[Index].num=1;edge[Index].next=pre[a];pre[a]=Index++;}}void Tarjan(int u,int father){int v,vv;for(v=pre[u];v!=-1;v=edge[v].next){vv=edge[v].to;if(!visit[vv]){low[vv]=dnf[vv]=++high;visit[vv]=1;Tarjan(vv,u);low[u]=MIN(low[u],low[vv]);if(low[vv]>dnf[u]&&edge[v].num==1)  //low[vv]>dnf[u]是桥,low[vv]>=dnf[u]是割点{ans[ansn++]=edge[v].m;}}else if(vv!=father)                //**{low[u]=MIN(low[u],dnf[vv]);}}}int main(){int m,i,t,a,b;scanf("%d",&t);while(t--){Index=0;memset(pre,-1,sizeof(pre));memset(visit,0,sizeof(visit));memset(edge,0,sizeof(edge));scanf("%d%d",&n,&m);for(i=0;i<m;i++){scanf("%d%d",&a,&b);Add_edge(a,b,i+1);Add_edge(b,a,i+1);}low[1]=dnf[1]=visit[1]=high=1;  //初始化ansn=0;Tarjan(1,-1);sort(ans,ans+ansn);             //序号排序输出printf("%d\n",ansn);for(i=0;i<ansn-1;i++){printf("%d ",ans[i]);}if(ansn)    printf("%d\n",ans[ansn-1]);if(t)puts("");}return 0;}