Asteroids(POJ 匈牙利算法）

最新推荐文章于 2020-03-04 22:28:03 发布

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最新推荐文章于 2020-03-04 22:28:03 发布

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分类专栏：算法文章标签：二分图的最大匹配

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本文介绍了一个基于二分图最大匹配的问题——如何使用最少次数的武器射击来清除网格中所有的小行星，并提供了完整的C++代码实现。

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Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23803		Accepted: 12909

Description

Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid.

Fortunately, Bessie has a powerful weapon that can vaporize all the asteroids in any given row or column of the grid with a single shot.This weapon is quite expensive, so she wishes to use it sparingly.Given the location of all the asteroids in the field, find the minimum number of shots Bessie needs to fire to eliminate all of the asteroids.

Input

* Line 1: Two integers N and K, separated by a single space.
* Lines 2..K+1: Each line contains two space-separated integers R and C (1 <= R, C <= N) denoting the row and column coordinates of an asteroid, respectively.

Output

* Line 1: The integer representing the minimum number of times Bessie must shoot.

Sample Input

Sample Output

Hint

INPUT DETAILS:
The following diagram represents the data, where "X" is an asteroid and "." is empty space:
X.X .X. .X.

OUTPUT DETAILS:
Bessie may fire across row 1 to destroy the asteroids at (1,1) and (1,3), and then she may fire down column 2 to destroy the asteroids at (2,2) and (3,2).

Source

USACO 2005 November Gold

二分图最大匹配的König定理证明尚不是很理解，只知道结论：最小点覆盖数=最大匹配数。
一个通俗易懂的讲解匈牙利算法的博文：http://blog.youkuaiyun.com/dark_scope/article/details/8880547

#include<iostream>
#include<string>
using namespace std;
const int maxn = 505;
int map[maxn][maxn];//可达矩阵存数据
int match[maxn];//顶点集V2中的点（下标）所匹配的V1中的点（值）
int visited[maxn];//记录是否已搜索
int n,v1,v2;//顶点数与顶点集
bool dfs(int x)
{
	int i, j;
	for ( i = 1; i <= v2; i++)//扫描V2
	{
		if (map[x][i] && visited[i]==0)//有边且未被搜索
		{
			visited[i] = 1;
			if (match[i]==0 || dfs(match[i]))//未匹配或匹配可更改
			{
				match[i] = x;//记录匹配的点
				return true;
			}
		}
	}
	return false;
}
int solve(int n,int result)
{
	for (int i = 1; i <= v1; i++)//逐个尝试v1中的点是否能被匹配
	{
		memset(visited, 0, sizeof(visited));//清空搜索标记
		if (dfs(i)) result++;
	}
	return result;//最大匹配数
}
int main()
{
	
	int  k;
	int x, y;
	while (cin>>n>>k)
	{
		memset(map, 0, sizeof(map));
		memset(match, 0, sizeof(match));
		v1 = v2 = n;
		while (k--)
		{
			cin >> x >> y;
			map[x][y] = 1;
		}
		cout<<solve(n,0)<<endl;
	}

}

转载地址：http://blog.youkuaiyun.com/yeruby/article/details/39315713
参考地址： 優YoU http://user.qzone.qq.com/289065406/blog/1299322465