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 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