poj-2536-Gopher II

最新推荐文章于 2018-11-27 00:02:33 发布

转载最新推荐文章于 2018-11-27 00:02:33 发布 · 124 阅读

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原文链接：http://www.cnblogs.com/zhang-yd/p/7989028.html

本文解析了POJ 2536 Gopher II问题，这是一个典型的二分图匹配问题，通过匈牙利算法实现解决方案。文章详细介绍了如何根据每个地鼠到洞穴的距离和时间限制来确定哪些地鼠可以安全到达洞穴。

poj-2536-Gopher II

Gopher II

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9268		Accepted: 3847

Description

The gopher family, having averted the canine threat, must face a new predator.

The are n gophers and m gopher holes, each at distinct (x, y) coordinates. A hawk arrives and if a gopher does not reach a hole in s seconds it is vulnerable to being eaten. A hole can save at most one gopher. All the gophers run at the same velocity v. The gopher family needs an escape strategy that minimizes the number of vulnerable gophers.

Input

The input contains several cases. The first line of each case contains four positive integers less than 100: n, m, s, and v. The next n lines give the coordinates of the gophers; the following m lines give the coordinates of the gopher holes. All distances are in metres; all times are in seconds; all velocities are in metres per second.

Output

Output consists of a single line for each case, giving the number of vulnerable gophers.

Sample Input

2 2 5 10
1.0 1.0
2.0 2.0
100.0 100.0
20.0 20.0

Sample Output

Source

Waterloo local 2001.01.27

17922621

2536

Accepted

448K

32MS

G++

1179B

2017-12-05 19:47:12

二分图匹配问题。

使用匈牙利算法解决。

/// poj-2536  

#include <cstdio>  
#include <cstring> 
const int MAXN = 100 + 10; 

struct Node{
	double x, y; 
}; 

Node gopher[MAXN], hole[MAXN]; 

int n, m, s, v, num[MAXN], gh[MAXN][MAXN], match[MAXN], vis[MAXN]; 

double DistSqu(const Node &a, const Node &b){
	return ( (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) ); 
}

bool dfs(int x){
	for(int i=0; i<num[x]; ++i){
		int y = gh[x][i]; 
		if(vis[y] == 0){
			vis[y] = 1; 
			if(match[y] < 0 || dfs(match[y])){
				match[y] = x; 
				return true; 
			}
		}
	}
	return false; 
}

int main(){
	freopen("in.txt", "r", stdin);  

	int ans; 
	while(scanf("%d %d %d %d", &n, &m, &s, &v) != EOF){
		for(int i=0; i<n; ++i){
			scanf("%lf %lf", &gopher[i].x, &gopher[i].y); 
		} 
		for(int i=0; i<m; ++i){
			scanf("%lf %lf", &hole[i].x, &hole[i].y); 
		} 
		int max_len_sq = s*s*v*v; 

		for(int i=0; i<n; ++i){
			num[i] = 0; 
			for(int j=0; j<m; ++j){
				if(DistSqu(gopher[i], hole[j]) <= max_len_sq ){
					gh[i][ num[i] ] = j; 
					++num[i]; 
				} 
			}
		} 
		memset(match, -1, sizeof(match)); 
		ans = 0; 
		for(int i=0; i<n; ++i){
			memset(vis, 0, sizeof(vis)); 
			if(dfs(i)){
				++ans; 
			}
		}
		printf("%d\n", n - ans );
	} 
	return 0; 
}

转载于:https://www.cnblogs.com/zhang-yd/p/7989028.html