题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1588
binshen的板子：
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int INF = 0x3f3f3f3f;
/*
* 求 无向图的割点和桥
* 可以找出割点和桥，求删掉每个点后增加的连通块。
* 需要注意重边的处理，可以先用矩阵存，再转邻接表，或者进行判重
*/
const int MAXN = 10010;
const int MAXM = 2000010;
struct Edge
{
int to,next;
int w;
bool cut;//是否为桥的标记
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN];
int Index,top;
bool Instack[MAXN];
bool cut[MAXN];
int add_block[MAXN];//删除一个点后增加的连通块
int bridge;
void addedge(int u,int v,int w)
{
edge[tot].to = v;edge[tot].next = head[u];edge[tot].cut = false;
edge[tot].w = w;
head[u] = tot++;
edge[tot].to = u;edge[tot].next = head[v];edge[tot].cut = false;
edge[tot].w = w;
head[v] = tot++;
}
void Tarjan(int u,int pre)
{
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
int son = 0;
int pre_num = 0;
for(int i = head[u];i != -1;i = edge[i].next)
{
v = edge[i].to;
if(v == pre && pre_num == 0){pre_num++;continue;}
if( !DFN[v] )
{
son++;
Tarjan(v,u);
if(Low[u] > Low[v])Low[u] = Low[v];
//桥
//一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。
if(Low[v] > DFN[u])
{
bridge++;
edge[i].cut = true;
edge[i^1].cut = true;
}
//割点
//一个顶点u是割点，当且仅当满足(1)或(2) (1) u为树根，且u有多于一个子树。
//(2) u不为树根，且满足存在(u,v)为树枝边(或称父子边，
//即u为v在搜索树中的父亲)，使得DFS(u)<=Low(v)
if(u != pre && Low[v] >= DFN[u])//不是树根
{
cut[u] = true;
add_block[u]++;
}
}
else if( Low[u] > DFN[v])
Low[u] = DFN[v];
}
//树根，分支数大于1
if(u == pre && son > 1)cut[u] = true;
if(u == pre)add_block[u] = son - 1;
Instack[u] = false;
top--;
}
void solve(int N)
{
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
memset(add_block,0,sizeof(add_block));
memset(cut,false,sizeof(cut));
Index = top = 0;
bridge = 0;
for(int i = 1;i <= N;i++)if(!DFN[i])Tarjan(i,i);
}
void init(){ tot = 0; memset(head,-1,sizeof(head));}
vector<int>G;
int n, m;
int main(){
int u, v, i, j, T; scanf("%d",&T);
while(T--){
scanf("%d %d",&n,&m);
init();
while(m--){
scanf("%d %d",&u,&v);
addedge(u,v,1);
}
solve(n);
G.clear();
for(i=0;i<tot;i+=2)if(edge[i].cut)G.push_back(i/2);
printf("%d\n",G.size());
for(i=0;i<G.size();i++){
printf("%d",G[i]+1);
i==G.size()-1?puts(""):printf(" ");
}
if(T)puts("");
}
return 0;
}
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;
#define N 10050
#define M 200005
int n,m;//n个点 m条边
struct Edge{
int from,to,val,nex;
bool cut;//记录这条边是否为割边
}edge[2*M];//双向边则 edge[i]与edge[i^1]是2条反向边
int head[N],edgenum;//在一开始就要 memset(head,-1,sizeof(head)); edgenum=0;
int low[N],dfn[N],tarjin_time;
void add(int u,int v,int w){
Edge E={u,v,w,head[u],0};
edge[edgenum]=E;
head[u]=edgenum++;
Edge E2={v,u,w,head[v],0};
edge[edgenum]=E2;
head[v]=edgenum++;
}
void tarjin(int u,int pre)
{
low[u]=dfn[u]= ++tarjin_time;
int flag=1;//flag是阻止双向边的反向边 i和i^1
for(int i=head[u];i!=-1;i=edge[i].nex)
{
int v=edge[i].to;
if(flag&&v==pre)
{
flag=0;
continue;
}
if(!dfn[v])
{
tarjin(v,u);
if(low[u]>low[v])low[u]=low[v];
if(low[v]>dfn[u])//是桥low[v]表示v能走到的最早祖先 有重边且u是v的最早祖先 则low[v]==dfn[u],所以不会当作桥
edge[i].cut=edge[i^1].cut=true;
}
else if(low[u]>dfn[v])low[u]=dfn[v];
}
}
void find_edgecut()
{
memset(dfn,0,sizeof(dfn));
tarjin_time=0;
for(int i=1;i<=n;i++)if(!dfn[i])tarjin(i,i);
}
void init(){memset(head, -1, sizeof head); edgenum = 0;}
vector<int>G;
int main(){
int u, v, i, j, T; scanf("%d",&T);
while(T--){
scanf("%d %d",&n,&m);
init();
while(m--){
scanf("%d %d",&u,&v);
add(u,v,1);
}
find_edgecut();
G.clear();
for(i=0;i<edgenum;i+=2)if(edge[i].cut)G.push_back(i/2);
printf("%d\n",G.size());
for(i=0;i<G.size();i++){
printf("%d",G[i]+1);
i==G.size()-1?puts(""):printf(" ");
}
if(T)puts("");
}
return 0;
}ZOJ 2588 Burning Bridges 求无向图桥 边双连通裸题,码迷,mamicode.com
ZOJ 2588 Burning Bridges 求无向图桥 边双连通裸题
原文地址：http://blog.csdn.net/acmmmm/article/details/24646745
