二分图匹配求最小边覆盖 建图方法中的黑白染色法,题目中说信号可以覆盖相邻两个块,那么我们可以将给定的地图染成国际象棋棋盘的样子,一个黑格可以与周围的四个白格共用信号,对于城市,从每一个黑格出发,向其周围的白格连边,那么这就是一个二分图,我们的把城市抽象成了点,所以我们的目的是找到最少的边覆盖所有的点,即最小边覆盖. 在二分图中,最小边覆盖=顶点数-最大匹配数. #include <iostream> #include <cstdio> #include <cstring>…

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…

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…

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…

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…

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: 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…

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…

http://poj.org/problem?id=3020 #include <cstdio> #include <cstring> #include <vector> using namespace std; ][]; ][]; ][]={,,,-,,,-,}; int n,m; vector <]; ]; void addedge(int from,int to){ e[; G[from].push_back(to); } int dfs(int s){ v…