ZOJ 2588 Burning Bridges 求无向图桥 边双连通裸题

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 200005int 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^1for(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;}