【Asteroids 】【POJ - 3041】（网络流-二分匹配-匈牙利）（思维）

最新推荐文章于 2019-02-02 20:10:29 发布

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最新推荐文章于 2019-02-02 20:10:29 发布

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分类专栏： acm训练

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在一片危险的陨石区中，Bessie必须使用她的武器清除所有障碍。通过二分匹配与匈牙利算法，我们找到了消除所有陨石所需的最少射击次数。

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题目：

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).

解题报告：你绝对不会想到的二分匹配。转换思维，将每个炸弹看作是一条边，而行列的话，是作为二分的点集。，只要是两个点被一个炸弹安排了，那么行列均没了，即求解最小覆盖。（利用匈牙利算法求解即可）

ac代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;

const int MAXN=1000;
int uN,vN;  //u,v数目
int g[MAXN][MAXN];//编号是0~n-1的 
int linker[MAXN];
bool used[MAXN];
bool dfs(int u)
{
    int v;
    for(v=1;v<=vN;v++)
        if(g[u][v]&&!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }    
        }  
    return false;  
}    
int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=1;u<=uN;u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    } 
    return res;   
}

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	uN=vN=n;
	while(m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		g[a][b]=1;
	}
	printf("%d\n",hungary());
}