POJ-3041-Asteroids

Asteroids

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 21095
Accepted: 11452

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).

Source

USACO 2005 November Gold

这是个较为典型的二分图匹配的问题，这个算法我也是今天才接触的感慨下，ACM的路要走还有好远，不过不也是不会的多才需要 更多的学习吗

对于二分图匹配的匈牙利算法我是在一个博客上看到学会的

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>int Map[510][510]（记录有关的边）,used[510]（是否匹配失败过）,match[510]（和谁匹配）;int N,C;bool Find(int x){for(int j = 1; j <= N; j++){if(Map[x][j] && !used[j]){used[j] = 1;if(match[j] == 0 || Find(match[j]))//当前没有匹配或则可以腾出来{match[j] = x;//和x匹配return true;}}}return false;}int main(){int u,v,Match;while(~scanf("%d %d",&N,&C)){memset(Map,0,sizeof(Map));for(int i = 1; i <= C; i++){scanf("%d %d",&u,&v);Map[u][v] = 1;}Match = 0;//记录最大的匹配数for(int i = 1; i <= N; i++){memset(used,0,sizeof(used));if(Find(i))Match++;}printf("%d\n",Match);}return 0;}