POJ 3041 Asteroids (二分匹配)

最新推荐文章于 2017-10-14 14:42:34 发布

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最新推荐文章于 2017-10-14 14:42:34 发布

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原文链接：http://www.cnblogs.com/wangdongkai/p/5643984.html

通过最大匹配算法解决在N*N网格中消除所有小行星的问题，利用行列作为点进行匹配，找到最少射击次数。

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Asteroids

Time Limit: 1000MS Memory Limit: 65536K

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).

题意：有一个N*N的格子，有些格子上有东西，你每次可以将一行或一列的东西打碎，问最少要几次才能全部打碎。

分析：直接以行和列为点进行最大匹配就行了。

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

const int MAXN = 600;
int uN,vN;
int G[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];

bool dfs(int u)
{
    for(int 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;
    memset(linker,-1,sizeof(linker));
    for(int u=1;u<=uN;u++){
        memset(used,false,sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}

int main()
{
    int N,K;
    scanf("%d%d",&N,&K);
    uN=vN=N;
    memset(G,0,sizeof(G));
    for(int i=1;i<=K;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        G[x][y]=1;
    }
    printf("%d\n",Hungary());

    return 0;
}