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

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).

Source

USACO 2005 November Gold

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题目来源 PKU 3041

匈牙利算法。

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#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int map[1000][1000];
int bmask[1000];
int ny,nx;
int cy[1000],cx[1000];
int n,k,a,b;
int xm,ym;
int findpath(int u)//找增广路径
 {
     int x,y;
     for(int i=1;i<=ny;i++)
      {
          if(map[u][i]&&bmask[i]==0)
           {
               bmask[i]=1;
               if(cy[i]==-1||findpath(cy[i]))
                 {
                     cy[i]=u;
                     cx[u]=i;
                     return 1;
                 }

           }
      }
      return 0;
 }
 int match()//算最大匹配
 {
     int res(0);
     for(int i=1;i<=nx;i++)
       cx[i]=-1;
     for(int j=1;j<=ny;j++)
       cy[j]=-1;
    for(a=1;a<=nx;a++)
     {

         if(cx[a]==-1)
          {
              for(b=1;b<=ny;b++)
                 bmask[b]=0;
              res+=findpath(a);

          }

     }
    return res;
 }
 int main()
 {
     cin>>n>>k;
     nx=n;
     ny=n;
     memset(map,0,sizeof(map));
     for(int i=1;i<=k;i++)
      {
          cin>>xm>>ym;
          map[xm][ym]=1;
      }
    cout<<match()<<endl;
    return 0;
 }