POJ3041-Asteroids

AsteroidsCrawling in process...

Crawling failedTime Limit:1000MS    Memory Limit:65536KB     64bit IO Format:%lld & %llu

SubmitStatus

Description

Input

Output

Sample Input

Sample Output

Hint

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).

题意：在N*N矩阵中存在K个障碍点，每次消除一行或者一列，求最少消除次数

题解：最大匹配-最小点覆盖-二分图匹配

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int map[500][500];//标记障碍点int pei[500];//匹配的列int vis[500];//标记访问过的行 int n,k;int dfs(int i){for(int j=1;j<=n;j++)//列循环{if(map[i][j]&&vis[j]==0){vis[j]=1;if(!pei[j]||dfs(pei[j])){pei[j]=i;return 1;}} }  return 0;}int main(){int a,b;while(scanf("%d%d",&n,&k)!=EOF){memset(map,0,sizeof(map));for(int i=0;i<k;i++){scanf("%d%d",&a,&b);map[a][b]=1; }memset(pei,0,sizeof(pei));int ans=0;for(int i=1;i<=n;i++){memset(vis,0,sizeof(vis));ans+=dfs(i);}printf("%d\n",ans);}return 0; }