Network of Schools A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that if B is in t…

题意: 求一个有向图中: (1)要选几个点才能把的点走遍 (2)要添加多少条边使得整个图强联通 分析: 对于问题1, 我们只要求出缩点后的图有多少个入度为0的scc就好, 因为有入度的scc可以从其他地方到达. 对于问题2, 每个入度为0的scc, 都可以补一条边可以变成强连通图, 每个出度为0的scc, 也可以补一条边使其变成强连通图. 所以答案就是max(入度为0scc个数,出度为0scc个数). #include<cstdio> #include<iostream> #inc…

题目链接:http://poj.org/problem?id=1236 Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19859   Accepted: 7822 Description A number of schools are connected to a computer network. Agreements have been developed among those…

Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16806   Accepted: 6643 Description A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a li…

Description A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the "receiving schools"). Note that if B is in the…

Network of Schools Description A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that…

Network of Schools Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13801   Accepted: 5505 Description A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a li…

一些学校连接到计算机网络.这些学校之间已经达成了协议: 每所学校都有一份分发软件的学校名单("接收学校"). 请注意,如果B在学校A的分发名单中,则A不一定出现在学校B的名单中您需要编写一个计划,计算必须接收新软件副本的最少学校数量, 以便软件根据协议(子任务A)到达网络中的所有学校.作为进一步的任务, 我们希望确保通过将新软件的副本发送到任意学校,该软件将覆盖网络中的所有学校. 为了实现这一目标,我们可能需要扩大新成员的接收者名单. 计算必须做的扩展的最小数目,以便我们发送新软件的任…

POJ 1236 Network of Schools(强连通 Tarjan+缩点) ACM 题目地址:POJ 1236 题意:  给定一张有向图,问最少选择几个点能遍历全图,以及最少加入�几条边使得有向图成为一个强连通图. 分析:  跟HDU 2767 Proving Equivalences(题解)一样的题目,只是多了个问题,事实上转化成DAG后就不难考虑了,事实上仅仅要选择入度为0的点即可了. 代码: /* * Author: illuz <iilluzen[at]gmail.com>…

POJ 1236 Network of Schools 题目链接 题意:题意本质上就是,给定一个有向图,问两个问题 1.从哪几个顶点出发,能走全全部点 2.最少连几条边,使得图强连通 思路: #include <cstdio> #include <cstring> #include <vector> #include <stack> using namespace std; const int N = 105; int n; vector<int>…