Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11378   Accepted: 5285 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 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12551   Accepted: 5771 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…

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…

poj1144 tarjan求割点 额,算法没什么好说的,只是这道题的读入非常恶心. 注意,当前点x是否是割点,与low[x]无关,只和low[son]和dfn[x]有关. 还有,默代码的时候记住分目标点是父亲还是孩子两种情况讨论. #include <cctype> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn=10…

Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11914   Accepted: 5519 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…

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…

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…

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: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 <…

解题关键:割点模板题. #include<cstdio> #include<cstring> #include<vector> #include<stack> using namespace std; #define N 1010 int n,m,ans,pd,son,cut[N],low[N],dfn[N]; stack<int>s; ; struct Edge{ int nxt; int to; int w; }e[maxn]; int he…