poj 3041 Asteroids （匈牙利算法,最小覆盖数 = 最大匹配数）

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poj 3041 Asteroids

　　Time Limit: 1 Sec
　　Memory Limit: 64 MB

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###

　　2

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).

题目地址：codevs 1022 覆盖

题目大意： 题目很简洁了：）

题解：

　　最小覆盖数 = 最大匹配数.

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AC代码

#include <stdio.h>
#include <cstring>
using namespace std;
const int N=505;
int n,m,Ans;
struct edge{
    int to,next;
}e[10005];
int cnt_edge,last[N];
void add_edge(int u,int v){
    e[++cnt_edge]=(edge){v,last[u]};last[u]=cnt_edge;
}
int res[N];
bool used[N];
bool find(int u){
    for(int i=last[u];i;i=e[i].next){
        int v=e[i].to;
        if(!used[v]){
            used[v]=1;
            if(!res[v] || find(res[v])){
                res[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        add_edge(x,y);
    }
    for(int i=1;i<=n;i++){
        memset(used,0,sizeof(used));
        if(find(i))Ans++;
    }
    printf("%d\n",Ans);
    return 0;
}

　　作者：skl_win
　　出处：https://www.cnblogs.com/shaokele/
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