题目连接:problemId=542" target="_blank">ZOJ 1542 POJ 1861 Network 网络 Network Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge Andrew is working as system administrator and is planning to establish a new network in his com…

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 14021   Accepted: 5484   Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the c…

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 14103   Accepted: 5528   Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the c…

Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: Accepted: Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can…

题意:有n个点,部分点之间可以连接无向边,每条可以连接的边都有一个权值.求一种连接方法将这些点连接成一个连通图,且所有连接了的边中权值最大的边权值最小. 解法:水题,直接用Kruskal算法做一遍就行了,不过还是应该仔细想想为什么Kruskal可行.原因是,在从小边往大边遍历的过程中(一直保持图为连通图),若判定某边i必须被连接,则因为图是连通图,所以连接比它小的边不可能使边i不需要连接,所以,要使边i不需要连接,必须连接比它大的边,根据题目要求,还是连接边i情况更优. tag:最小生成树 /*…

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 15268   Accepted: 5987   Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the c…

题意:求解最小生成树,以及最小瓶颈生成树上的瓶颈边. 思路:只是求最小生成树即可.瓶颈边就是生成树上权值最大的那条边. //#include <bits/stdc++.h> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> #define INF 0x7f7f7f7f #define pii pair<int,int> #defin…

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>…

poj 3417 Network(tarjan lca) 先给出一棵无根树,然后下面再给出m条边,把这m条边连上,然后每次你能毁掉两条边,规定一条是树边,一条是新边,问有多少种方案能使树断裂. 我们设添加了一条新边后,树形成了一个环,表示为x->y->lca(x,y),我们将其中的边都覆盖一次.添加了多条新边后,可知树上有些边是会被多次覆盖的,画图很容易发现,但一个树边被覆盖了2次或以上,它就是一条牢固的边,就是说毁掉它再毁掉任何一条新边都好,树都不会断裂.所以我们只要统计被覆盖过零次或一次的…