Asteroids(poj 3041,二分图最大匹配)

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本文介绍了一个基于网格的消除游戏算法，目标是最少次数地清除所有障碍物。通过将问题转化为行与列的最大匹配问题，利用图算法求解最小射击次数。

http://poj.org/problem?id=3041

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29327#problem/C

C - Asteroids

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

3 4

1 1

1 3

2 2

3 2

Sample Output

Hint

INPUT DETAILS:

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

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).

解析：

题意：

给出星星的位置，要求把星星全部消灭，每次只能消灭一整行或者一整列，

问最少需要消灭几次

思路：

可以转化为行与列的匹配，求出的最大匹配数就是答案

1184 KB 32 ms C++ 685 B

#include<string.h>
#include<stdio.h>
#include<algorithm>
#include <iostream>
using namespace std;
const int maxn=500+10;
int n;
int vis[maxn],result[maxn];
int map[maxn][maxn];
int find(int a)
{
	for(int i=1;i<=n;i++)
	{
		if(map[a][i]&&!vis[i])
		{
			vis[i]=1;
			if(result[i]==0||find(result[i]))
			{
				result[i]=a;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{    
    int i,k;
	while(scanf("%d%d",&n,&k)!=EOF)
	{int u,v;
	memset(map,0,sizeof(map));
	for(i=1;i<=k;i++)
	{
      scanf("%d%d",&u,&v);
      map[u][v]=1;
	}
	int ans=0;
	memset(result,0,sizeof(result));
	for(i=1;i<=n;i++)
	{
		memset(vis,0,sizeof(vis));
		if(find(i))
		ans++;
	}
	printf("%d\n",ans);
}
  return 0;
}