poj3041 Asteroids 最小点覆盖 二分图

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13927 Accepted: 7566

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 41 11 32 23 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).

  把每行和列分别看成两组，交点看成边，那么每个边至少有一个点被覆盖，这道题就是求最小点覆盖。

  最小点覆盖：用最少的点覆盖所有的边（每条边至少一个顶点被覆盖），最小点覆盖=边最大匹配。充分性：M个点是足够的，它们覆盖最大匹配的所有边后，若还有边没被覆盖，则可以得到更大匹配，矛盾。必要性：最大匹配两两无公共点，至少要M个点才能把这些边覆盖。

  最小边覆盖：最小路径覆盖就是找出最小的路径条数，使之成为P的一个路径覆盖。最小边覆盖=点数-最大匹配数。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<cstdlib>#define INF 0x3f3f3f3f#define MAXN 510#define MAXM 15#define MAXNODE 30000#define pii pair<int,int>using namespace std;int N,K,vis[MAXN],match[MAXN],w[MAXN][MAXN];int augment_path(int u){    for(int i=1;i<=N;i++) if(w[u][i]&&!vis[i]){        vis[i]=1;        if(!match[i]||augment_path(match[i])){            match[i]=u;            return 1;        }    }    return 0;}int hungary(){    memset(match,0,sizeof(match));    int ret=0;    for(int i=1;i<=N;i++){        memset(vis,0,sizeof(vis));        if(augment_path(i)) ret++;    }    return ret;}int main(){    while(scanf("%d%d",&N,&K)!=EOF){        memset(w,0,sizeof(w));        int u,v;        for(int i=0;i<K;i++){            scanf("%d%d",&u,&v);            w[u][v]=1;        }        printf("%d\n",hungary());    }    return 0;}