匈牙利模板--poj 3041最小匹配

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23596 Accepted: 12806

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).

这道题只用套用匈牙利算法就可以解决 

将问题看成匹配 题就可以转换为匈牙利算法了

#include<iostream>#include <string.h>#define maxn 501using namespace std;int n,k;bool map[maxn][maxn],vis[maxn];int  connect[maxn];void input(){int i;    for (i =0; i < k; i++){        int a, b;        cin>>a>>b;        map[a][b] =1;    }}bool work(int a){int i;    for (i =1; i <= n; i++)        if (map[a][i] &&!vis[i]){            vis[i] =true;            if (connect[i] ==-1|| work(connect[i]))            {                connect[i] = a;                return true;            }        }    return false;}int main(){ios::sync_with_stdio(false);  //没有加速反而多了十几ms    cin>>n>>k;    input();    int i;    memset(connect, -1, sizeof(connect));int ans =0;    for (i =1; i <= n; i++){        memset(vis, false, sizeof(vis));        if(work(i)) ans++;    }    cout<<ans;    return 0;}