Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6445   Accepted: 3182 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…

题目传送门 /* 题意:*的点占据后能顺带占据四个方向的一个*,问最少要占据多少个 匈牙利算法:按坐标奇偶性把*分为两个集合,那么除了匹配的其中一方是顺带占据外,其他都要占据 */ #include <cstdio> #include <algorithm> #include <cstring> #include <vector> using namespace std; ; const int INF = 0x3f3f3f3f; ][]; ][]; boo…

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…

传送门:http://poj.org/problem?id=3020 Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11098   Accepted: 5464 Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobi…

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…

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…

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[…

Antenna Placement Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5645 Accepted: 2825 Description The Global Aerial Research Centre has been allotted the task of building the fifth generation of mobile phone nets in Sweden. The most striki…

Antenna Placement Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6438   Accepted: 3176 看了别人的题解才过的... 渣啊.. 最基本的是构图 城市才是要构造的二分图的顶点! 构造方法例如以下: 比如输入: *oo *** O*o 时,能够抽象为一个数字地图: 100 234 050 数字就是依据输入的城市次序作为该城市的编号,0代表该位置没有城市. 然后依据题目的"范围"规…

题目地址:http://poj.org/problem?id=3020 输入一个字符矩阵,'*'可行,'o'不可行.因为一个点可以和上下左右四个方向的一个可行点组成一个集合,所以对图进行黑白染色(每个点的值为其横纵坐标之和),然后就可划分为二分图,进行最大匹配.最后最大匹配数加剩下的单个点数量即为所求. #include<cstdio> #include<iostream> #include<string.h> #include<algorithm> #in…