POJ3041-Asteroids

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21806		Accepted: 11839

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 4
1 1
1 3
2 2
3 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：有一个n*n的矩阵,在矩阵上有m个行星,每一个武器都可以消灭一行或者一列的行星，要求使用最小的武器消灭所有的行星

解题思路：二分图，将每一行看成一个节点,分到一边,每一列也看成一个节点,分到另一边,如果每行和每列有交点就联边，要使用最少的武器消灭所有的行星,就是求最大匹配的数

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int n,m;
int x[525],y[525];
int s[525],nt[20050],e[20050];
int visit[525];

bool path(int k)
{
    for(int i=s[k];~i;i=nt[i])
    {
        int ee=e[i];
        if(!visit[ee])
        {
            visit[ee]=1;
            if(y[ee]==-1||path(y[ee]))
            {
                y[ee]=k;
                x[k]=ee;
                return 1;
            }
        }
    }
    return 0;
}

void MaxMatch()
{
    int ans=0;
    memset(x,-1,sizeof x);
    memset(y,-1,sizeof y);
    for(int i=1;i<=n;i++)
    {
        if(x[i]==-1)
        {
            memset(visit,0,sizeof visit);
            if(path(i)) ans++;
        }
    }
    printf("%d\n",ans);
}

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        int cnt=1,ee,ss;
        memset(s,-1,sizeof s);
        memset(nt,-1,sizeof nt);
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&ss,&ee);
            nt[cnt]=s[ss],s[ss]=cnt,e[cnt++]=ee;
        }
        MaxMatch();
    }
    return 0;
}