POJ3041 二分图（性质）最小点覆盖等于最大匹配数（匈牙利模板题）

二分图的性质

二分图中，点覆盖数是匹配数。
    (1) 二分图的最大匹配数等于最小覆盖数，即求最少的点使得每条边都至少和其中的一个点相关联，很显然直接取最大匹配的一段节点即可。
    (2) 二分图的独立数等于顶点数减去最大匹配数，很显然的把最大匹配两端的点都从顶点集中去掉这个时候剩余的点是独立集，这是|V|-2*|M|，同时必然可以从每条匹配边的两端取一个点加入独立集并且保持其独立集性质。
    (3) DAG的最小路径覆盖，将每个点拆点后作最大匹配，结果为n-m，求具体路径的时候顺着匹配边走就可以，匹配边i→j',j→k',k→l'....构成一条有向路径。

     （4）最大匹配数=左边匹配点+右边未匹配点。因为在最大匹配集中的任意一条边，如果他的左边没标记，右边被标记了，那么我们就可找到一条新的增广路，所以每一条边都至少被一个点覆盖。

     （5）最小边覆盖=图中点的个数-最大匹配数=最大独立集。

题目：http://poj.org/problem?id=3041

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23262 Accepted: 12611

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：给出n*n矩阵，并且给出障碍物坐标，让你求最小的武器使用次数消灭所有障碍。

思路：以行列分别代表一个点集，形成二分图，在利用最小点覆盖在二分图中等于最大匹配数套用匈牙利模板。

Code：

#include <iostream>#include <algorithm>#include <string.h>using namespace std;const int AX = 500+66;int n,k;int g[AX][AX];int used[AX];int linker[AX];bool dfs( int u ){int v;for( v = 1 ; v <= n ; v++ ){if( !used[v] && g[u][v] ){used[v] = 1;if( linker[v] == -1 || dfs(linker[v]) ){linker[v] = u;return true;}}}return false;}int xyl(){int ans = 0;int u;memset(linker,-1,sizeof(linker));for( u = 1 ; u <= n ;u ++ ){memset(used,0,sizeof(used));if( dfs(u) ) ans ++;}return ans;}int main(){ios_base::sync_with_stdio(false);cin.tie(0);cin>>n>>k;int x,y;for( int i = 0 ; i < k ; i++ ){cin>>x>>y;g[x][y] = 1;}cout<<xyl()<<endl;return 0;}