学长写的: #include<cstdio>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#include<cstring>#include<vector>using namespace std;#define maxn 10005int dfn[maxn];///代表最先遍历到这个点的时间int low[maxn];///这个点…

<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…

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…

传送门 题意: 有一张联通网络,求出所有的割点: 对于割点 u ,求将 u 删去后,此图有多少个联通子网络: 对于含有割点的,按升序输出: 题解: DFS求割点入门题,不会的戳这里…

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…

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…

输入数据处理正确其余的就是套强联通的模板了 #include <iostream> #include <cstdlib> #include <cstdio> #include <algorithm> #include <vector> #include <queue> #include <cmath> #include <stack> #include <cstring> using namespa…

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>…

    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…

SPF 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 failure of a single node, 3, in the network on the left would prevent some of the s…