POJ3041：Asteroids（二分图匹配）

Asteroids

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23141 Accepted: 12549

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

题意：N*N地图，K个位置有炸弹，每次可以将一行或一列炸弹消除，最少几次消除全部炸弹。

思路：每个炸弹的XY看成一条边，就是二分图的最小点覆盖问题了，即求最大匹配。

# include <iostream># include <cstdio># include <cstring># include <vector>using namespace std;int n, k, m[503][503], vis[503], l[503];vector<int>v[503];int dfs(int u){    for(int i=0; i<v[u].size(); ++i)    {        int j = v[u][i];        if(!vis[j])        {            vis[j] = 1;            if(l[j] == -1 || dfs(l[j]))            {                l[j] = u;                return 1;            }        }    }    return 0;}int main(){    while(~scanf("%d%d",&n,&k))    {        memset(l, -1, sizeof(l));        for(int i=1; i<=n; ++i) v[i].clear();        for(int i=0; i<k; ++i)        {            int x, y;            scanf("%d%d",&x,&y);            v[x].push_back(y);        }        int ans = 0;        for(int i=1; i<=n; ++i)        {            memset(vis, 0, sizeof(vis));            if(dfs(i)) ++ans;        }        printf("%d\n",ans);    }    return 0;}