Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7788   Accepted: 3880 Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most st…

链接:poj 3020 题意:一个矩形中,有n个城市'*'.'o'表示空地,如今这n个城市都要覆盖无线,若放置一个基站, 那么它至多能够覆盖本身和相邻的一个城市,求至少放置多少个基站才干使得全部的城市都覆盖无线? 思路:求二分图的最小路径覆盖(无向图) 最小路径覆盖=点数-最大匹配数 注:由于为无向图,每一个顶点被算了两次,最大匹配为原本的两倍. 因此此时最小路径覆盖=点数-最大匹配数/2 #include<stdio.h> #include<string.h> int edge[…

http://poj.org/problem?id=3020 Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7565   Accepted: 3758 Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile ph…

链接: http://poj.org/problem?id=3020 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/M Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5500   Accepted: 2750 Description The Global Aerial Research Centr…

The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striking reason why they got the job, is their discovery of a new, highly noise resistant, antenna. It is called 4D…

Antenna Placement Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3020 Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone n…

原题传送:http://poj.org/problem?id=3020 Antenna Placement Time Limit: 1000MS Memory Limit: 65536K Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striking…

Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6334   Accepted: 3125 Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most st…

二分图题目 当时看到网上有人的博客写着最小边覆盖,也有人写最小路径覆盖,我就有点方了,斌哥(kuangbin)的博客上只给了代码,没有解释,但是现在我还是明白了,这是个最小路径覆盖(因为我现在还不知道啥叫最小边覆盖). 有一篇博客如下写道:最小路径覆盖只对有向无环图而言,且并不要求原图是二分图,给所有点一个分身,让他们分别处于两个集合就可以,求出的最小路径覆盖 = n - 最大匹配值. 证明:假设最大匹配值是0,那原先一共有n个路径,每次多一个匹配,这样的路径就减少1,证明完毕. 那么回到这个题…

题意:图没什么用  给出一个地图 地图上有 点 一次可以覆盖2个连续 的点( 左右 或者 上下表示连续)问最少几条边可以使得每个点都被覆盖 最小路径覆盖       最小路径覆盖=|G|-最大匹配数                   证明:https://blog.csdn.net/qq_34564984/article/details/52778763 证明总的来说就是尽可能多得连边 边越多 可以打包一起处理得点就越多(这里题中打包指连续得两个点只需要一条线段就能覆盖) 拆点思想   :匈牙…