Tarjan 割边(桥)

//Tarjan 割边
//当dfn[u]<low[v]时u与v之间的边为割边
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int n,m,cnt,ans,head[10001];
int dot,dfn[10001],low[10001];
int cut[200001];
struct uio{
    int from,to,next,id;
}edge[200001];
void add(int x,int y)
{
    edge[++cnt].next=head[x];
    edge[cnt].from=x;
    edge[cnt].to=y;
    edge[cnt].id=cnt;
    head[x]=cnt;
}
void tarjan(int x,int fa)
{
    dfn[x]=low[x]=++dot;
    for(int i=head[x];i;i=edge[i].next)
    {
        int y=edge[i].to;
        if(!dfn[y])
        {
            tarjan(y,x);
            low[x]=min(low[x],low[y]);
            if(low[y]>dfn[x])
                 cut[edge[i].id]=1;
        }
        else if(y!=fa)
            low[x]=min(low[x],dfn[y]);
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
        add(v,u);
    }
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i,i);
    for(int i=1;i<=cnt;i++)
        if(cut[i])
            ans++;
    printf("%d\n",ans);
    for(int i=1;i<=cnt;i++)
        if(cut[i])
            printf("%d %d\n",edge[i].from,edge[i].to);
     return 0;
}

原文地址:https://www.cnblogs.com/water-radish/p/9280524.html

时间: 2024-08-09 10:32:54

Tarjan 割边(桥)的相关文章

tarjan,割边,桥,割点

这里是tarjan的基础知识, 求割点和割边 先来求割边, #include <cstdio> #include <iostream> #include <bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned long long ull; #define ls (t<<1) #define rs ((t<<1)|1) #define mid ((l+r

HDU 4738 Caocao&#39;s Bridges tarjan求桥

Caocao's Bridges Problem Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn't give up. Caocao's army still was not good at water battles, so he came up with another idea. He built many islands in the Chan

HDU 4738——Caocao&#39;s Bridges——————【求割边/桥的最小权值】

Caocao's Bridges Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 4738 Description Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn't give up. Caocao's army st

Tarjan求桥和割点

//Tarjan 求桥和割点 Tarjan(u,fa) { DFN[u]=LoW[u]=++time; Cu=grey; for each e=(u,v) { Tarjan(v,u); if(Cv=white) { low[u]=min(low[u],low[v]); }else { low[u]=min(low[u],DFN[v]); } } }

【模板】【hdu4738】Caocao&#39;s Bridges——tarjan求桥

题目链接 题目大意: 曹操有N个岛,这些岛用M座桥连接起来,每座桥有士兵把守(也可能没有), 诸葛亮把所有炸弹都带走了,只留下一枚给周瑜(真狠). 周瑜想让这N个岛不连通,但需要派出不小于守桥士兵数的人去炸桥,因为只有一枚炸弹,因此只够炸掉一座桥. 分析: 很明显的求代价最小的桥,当然这道题有几个特殊的地方: 1.图本来就不联通,输出0: 2.无解(不存在桥),输出-1: 3.没人把守,但你还是得派一个人炸桥,输出1: 4.一般情况,输出最小代价. 剩下的就是模板了,不过需要注意的一点是,这道题

tarjan求桥、割顶

若low[v]>dfn[u],则(u,v)为割边.但是实际处理时我们并不这样判断,因为有的图上可能有重边,这样不好处理.我们记录每条边的标号(一条无向边拆成的两条有向边标号相同),记录每个点的父亲到它的边的标号,如果边(u,v)是v的父亲边,就不能用dfn[u]更新low[v].这样如果遍历完v的所有子节点后,发现low[v]=dfn[v],说明u的父亲边(u,v)为割边. void tarjan(int x) { vis[x]=1; dfn[x]=low[x]=++num; for(int i

hdu-4738(tarjan割边)

题意:给你n个点,m条边,边有权值,问你最小的花费使图不连通: 解题思路:就是求边权最小的割边,但这道题有坑点: 1.有重边(桥的两个点有重边时,你去掉一条边并没什么d用): 2.当权值为0的时候,我们也需要放一个人(被这个坑死了0.0): #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #define maxn 1100 #define inf 0x3f3f

[UVA796]Critical Links(割边, 桥)

题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=737 求桥的数量,也就是割边的数量.输入有点小坑,左右括号外必须得有个空格才行,起初以为是转义的问题. 1 /* 2 ━━━━━┒ギリギリ♂ eye! 3 ┓┏┓┏┓┃キリキリ♂ mind! 4 ┛┗┛┗┛┃\○/ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8

UVA796 - Critical Links(Tarjan求桥)

In a computer network a link L, which interconnects two servers, is considered critical if there are atleast two servers A and B such that all network interconnection paths between A and B pass through L.Removing a critical link generates two disjoin