ZOJ 1654 Place the Robots 二分图最大匹配

最新推荐文章于 2022-03-31 11:30:30 发布

weixin_30470857

最新推荐文章于 2022-03-31 11:30:30 发布

阅读量82

点赞数

CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/PotremZ/p/ZOJ1654.html

解决一个关于在地图上放置机器人的问题，地图由空地、草地和墙组成。目标是在满足一定条件的情况下，放置尽可能多的机器人，使得它们无法互相攻击。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

$ \rightarrow $ 戳我进ZOJ原题

Place the Robots

Time Limit: 5 Seconds $ \quad $ Memory Limit: 32768 KB

Robert is a famous engineer. One day he was given a task by his boss. The background of the task was the following: 　 Given a map consisting of square blocks. There were three kinds of blocks: Wall, Grass, and Empty. His boss wanted to place as many robots as possible in the map. Each robot held a laser weapon which could shoot to four directions (north, east, south, west) simultaneously. A robot had to stay at the block where it was initially placed all the time and to keep firing all the time. The laser beams certainly could pass the grid of Grass, but could not pass the grid of Wall. A robot could only be placed in an Empty block. Surely the boss would not want to see one robot hurting another. In other words, two robots must not be placed in one line (horizontally or vertically) unless there is a Wall between them. 　 Now that you are such a smart programmer and one of Robert's best friends, He is asking you to help him solving this problem. That is, given the description of a map, compute the maximum number of robots that can be placed in the map. 　

Input

The first line contains an integer $ T (\le 11) $ which is the number of test cases. 　 For each test case, the first line contains two integers $ m $ and $ n (1 \le m, n \le 50) $ which are the row and column sizes of the map. Then $ m $ lines follow, each contains $ n $ characters of '#', '*', or 'o' which represent Wall, Grass, and Empty, respectively. 　

Output

For each test case, first output the case number in one line, in the format: "Case :id" where id is the test case number, counting from 1. In the second line just output the maximum number of robots that can be placed in that map. 　

Sample Input

 2
 4 4
 o***
 *###
 oo#o
 ***o
 4 4
 #ooo
 o#oo
 oo#o
 ***#

Sample Output

 Case :1
 3 
 Case :2
 5

Author: XU, Luchuan

Source: ZOJ Monthly, October 2003 　

题目大意

有一个 $ N \times M (N,M \le 50) $ 的棋盘
棋盘的每一格是三种类型之一：空地、草地、墙。
机器人只能放在空地上。
在同一行或者同一列的两个机器人，若他们之间没有墙，则他们可以互相攻击。
问给定的棋盘，最多可以放置多少个机器人，使它们不能互相攻击。

题解

把每一行、每一列被墙隔开，并且包含空地的连续区域称为“块”。
显然，在一个块之中，最多只能放一个机器人，把这些块编上号。
把每个横向块看作左部的点，竖向块看作右部的点
若两个块有公共的空地，则在它们之间连边
求二分图的最大匹配，即可得到答案

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define N 55
#define M 2505
vector<int>e[M];
int T,n,m,ans;
bool f;
char mp[N][N];
int cntx,cnty,x[N][N],y[N][N]; 
int link[M]; bool vis[M];
bool dfs(int u){
	for(int i=0;i<e[u].size();++i)
		if(!vis[e[u][i]]){
			int v=e[u][i];
			vis[v]=1;
			if(!link[v]||dfs(link[v])){
				link[v]=u;
				return 1;
			}
		}
	return 0;
}
int main(){
	scanf("%d",&T);
	for(int cases=1;cases<=T;++cases){
		memset(x,0,sizeof(x)); 
		memset(y,0,sizeof(y));
		memset(link,0,sizeof(link));
		cntx=cnty=ans=0;
		scanf("%d %d",&n,&m);
		for(int i=1;i<=n;++i){
			f=0; scanf("%s",mp[i]+1);
			for(int j=1;j<=m;++j)
				if(mp[i][j]=='o'){
					if(!f) ++cntx; f=1;
					x[i][j]=cntx;
				}else if(mp[i][j]=='#') f=0;
		}
		for(int j=1;j<=m;++j){
			f=0;
			for(int i=1;i<=n;++i)
				if(mp[i][j]=='o'){
					if(!f) ++cnty; f=1;
					y[i][j]=cnty;
				}else if(mp[i][j]=='#') f=0;
		}
		for(int i=1;i<=cntx;++i) e[i].clear();
		for(int i=1;i<=n;++i)
			for(int j=1;j<=m;++j)
				if(x[i][j]&&y[i][j]) 
					e[x[i][j]].push_back(y[i][j]);
		for(int i=1;i<=cntx;++i){
			memset(vis,0,sizeof(vis)); 
			if(dfs(i)) ++ans;
		}
		printf("Case :%d\n%d\n",cases,ans);
	}
	return 0;
}

转载于:https://www.cnblogs.com/PotremZ/p/ZOJ1654.html