传送门:Network 题意:给出一张无向图,求割点的个数. 分析:模板裸题,直接上模板. #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <iostream> #include <algorithm> #include <queue> #include <cstdlib> #include <…

题目大意:有向图求割点 题目思路: 一个点u为割点时当且仅当满足两个两个条件之一: 1.该点为根节点且至少有两个子节点 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 dfn(u)<=low(v). 然后注意读入,很容易RE #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<vector&…

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=5&page=show_problem&problem=251  Network  A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbe…

题意抽象: 给定一个无向图,输出割点个数. 割点定义:删除该点后,原图变为多个连通块. 考虑一下怎么利用tarjan判定割点: 对于点u和他相连的当时还未搜到的点v,dfs后如果DFN[u]<=low[v],那么u是割点.(搜v得到的是一个不会倒卷回来的子图) 另外注意一下tarjan搜索时的起始点如果有多个儿子那么它也是割点. AC代码: #include<cstdio> #include<cstring> #define rep(i,a,b) for(int i=a;i&…

Network 题目链接:https://vjudge.net/problem/UVA-315 Description: A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N. No two places have the same number. The li…

题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=251 求割点,除了输入用strtok和sscanf处理输入以外,对于求割点的tarjan算法有了进一步理解. 特别注意88行,如果u是根并且至少两个儿子,那它一定是割点无误,还有第二个情况用如图代表: 这个例子里显然:low[4]=2,dfn[4]=4,dfn[3]=3.现dfs到…

链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=251 http://poj.org/problem?id=1144 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82833#problem/B 首先输入一个N(多实例,0结束),下面有不超过N行的数,每行的第一个数字代表…

题目链接:https://vjudge.net/problem/UVA-315 A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N. No two places have the same number. The lines are bidirectional…

本题大意:求一个无向图额割点的个数. 本题思路:建图之后打一遍模板. /************************************************************************* > File Name: uva-315.network.cpp > Author: CruelKing > Mail: 2016586625@qq.com > Created Time: 2019年09月06日 星期五 17时15分07秒 本题思路:就是求图中有多…

Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12707   Accepted: 5835 Description A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N…

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=251  Network  A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers…

Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17016   Accepted: 7635 Description A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N…

Network Description A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect…

转载请注明出处,谢谢:http://www.cnblogs.com/KirisameMarisa/p/4319585.html   ---by 墨染之樱花 [题目链接]http://poj.org/problem?id=1144 [题目描述](半天才看明白...)给图求割点个数 [思路]直接套求割点的模板即可,就是要注意输入比较坑.代码见下,附注释 #include <iostream> #include <ios> #include <iomanip> #includ…

<题目链接> 题目大意: 给出一个无向图,求出其中的割点数量. 解题分析: 无向图求割点模板题. 一个顶点u是割点,当且仅当满足 (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 dfn(u)<=low(v).(也就是说V没办法绕过 u 点到达比 u dfn要小的点) 注:这里所说的树是指,DFS下的搜索树. #include <cstdio> #include <cstring&g…

题目:http://poj.org/problem?id=1144 求割点.判断一个点是否是割点有两种判断情况: 如果u为割点,当且仅当满足下面的1条 1.如果u为树根,那么u必须有多于1棵子树 2.如果u不为树根,那么(u,v)为树枝边,当Low[v]>=DFN[u]时. 然后根据这两句来找割点就可以了. 模版题,就是题意看不懂.看了题解.这题算是废了,就当贴模版用吧. #include <iostream> #include <stdio.h> #include <…

题目地址:id=1144">POJ 1144 求割点.推断一个点是否是割点有两种推断情况: 假设u为割点,当且仅当满足以下的1条 1.假设u为树根,那么u必须有多于1棵子树 2.假设u不为树根.那么(u,v)为树枝边.当Low[v]>=DFN[u]时. 然后依据这两句来找割点就能够了. 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstri…

题目链接:https://vjudge.net/contest/67418#problem/B 题意:给一个无向连通图,求出割点的数量.首先输入一个N(多实例,0结束),下面有不超过N行的数,每行的第一个数字代表后面的都和它存在边,0表示行输入的结束. 题解:简单的求割点模版,所谓割点就是去掉这一个点还有于这个点链接的边之后使得原来的图连通块增加. 由于这是模版题代码会加上注释. #include <iostream> #include <cstring> using namesp…

网上的题解大都模糊,我可能写的也比较模糊吧,讲究看看. 大致题意: 原图没有一个割点时,特殊考虑,至少ans1=2个通风井,方案数n*(n-1)/2; 原图上有多个割点时,每个(由割点限制成几部分的)联通块个数即为ans1:需要dfs进行vis标记和iscut区分,不重不漏: ans2,建设时避开在割点上建设通风井(通风井数量可最小化,以免通风井损毁后还需再建一个以备万一):求解时:当一个颜色块有两个割点时,摧毁一个蚂蚁们总可以通过另一个割点紧急转移:当一个颜色块有仅一个割点时,摧毁割点后就必须…

题目链接:http://poj.org/problem?id=1144 题目大意:给以一个无向图,求割点数量. 这道题目的输入和我们一般见到的不太一样. 它首先输入 \(N\)(\(\lt 100\))表示点的数量(\(N=0\)表示文件输入结束). 然后接下来每行输入一组数字. 如果这一组数字只包含一个 \(0\) ,说明本组测试数据输入结束: 否则,假设这些数可以拆分成 \(a_1,a_2,a_3, \cdots ,a_m\),则说明 \(a_1\) 这个点到 \(a_2,a_3, \cdo…

SPF Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7678   Accepted: 3489 Description Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a…

题意:给出一个无向图,求割点以及去除这个点后图分为几部分: 思路:割点定义:去掉该点后图将分成几个部分.割点:(1)当k为根节点且有>1个分支,则去除该点后图便被分成几个分支.(2)DFN[v]<Low[j]表示v的子节点不会有回路回到v的祖先. 代码: #include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAXN 1005 #define MA…

poj_1144Network(tarjan求割点) 标签: tarjan 割点割边模板 题目链接 Network Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12356 Accepted: 5688 Description A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting se…

描述 A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N. No two places have the same number. The lines are bidirectional and always connect together two plac…

Description A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect togethe…

poj2117 Electricity Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 3603   Accepted: 1213 Description Blackouts and Dark Nights (also known as ACM++) is a company that provides electricity. The company owns several power plants, each of…

    A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N. No two places have the same number. The lines are bidirectional and always connect together two pla…

题目:http://poj.org/problem?id=1523 题目解析: 注意题目输入输入,防止PE,题目就是求割点,并问割点将这个连通图分成了几个子图,算是模版题吧. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <stack> #include <string> #define N 10010…

求割点 一种显然的n^2做法: 枚举每个点,去掉该点连出的边,然后判断整个图是否联通 用tarjan求割点: 分情况讨论 如果是root的话,其为割点当且仅当下方有两棵及以上的子树 其他情况 设当前节点为u,一个儿子节点为v 存在low[v]>=dfn[u],也就是说其儿子节点v能连到的最前面的点都在u的下面 也就是当u断开的时候,u之前的点与以v为根的子树必然分成两个独立的块 那么这个时候u就是割点 Network A Telephone Line Company (TLC) is estab…

题意: 给个无向图,问有多少个割点,对于每个割点求删除这个点之后会产生多少新的点双联通分量 题还是很果的 怎么求割点请参考tarjan无向图 关于能产生几个新的双联通分量,对于每个节点u来说,我们判断他是否是割点,即判断是否满足他的儿子v的low[v]>dfn[u] 而这个时候割掉这个点就会让双联通分量增加,所以搞一个数组记录一下这个操作的次数就行 请注意在是否是根节点的问题上特判 !!注意输出格式!! #include<cstdio> #include<algorithm>…