POJ 3041 Asteroids（二分图匹配）

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#POJ #二分图匹配

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解决一个给定的n*n网格中，通过消除行或列来清除所有行星的最少操作次数问题。采用图论中的二分图匹配算法实现。

题目链接：http://poj.org/problem?id=3041

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

Sample Output

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).

题目大意：给定n*n的格子，k个点的行星坐标x，y，每操作一次可以消灭一行或一列的行星，问最少操作次数。

题目思路：把输入的x，y用边连接起来，则所有的边相当于给定的点，就相当于操作某个二分图里面的某个点则可以把与该点链接的所有所有边抹掉。即最小点覆盖，性质：最小点覆盖=最大匹配数。

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 550;

//时间复杂度：O（N*E）
int line[N][N];  //i,j之间是否存在关系,即可以匹配,记得初始化
int used[N];     //每一轮寻找增广路径时第i个妹子是否被使用过
int Next[N];     //Next[i]=x代表第i个妹子和x号男生是匹配的,记得初始化
int n,m;

bool Find(int x)
{
    for(int i=1;i<=m;i++)
    {
        if(line[x][i]&&!used[i])
        {
            used[i] = 1;             //第i个妹子被使用了
            if(Next[i]==0||Find(Next[i]))   //判断第i个妹子是否有了男朋友，没有的话就可以匹配,有的话就把next[i]号男生转移掉再匹配
            {
                Next[i] = x;
                return true;
            }
        }
    }
    return false;
}

int maxMatch()
{
    memset(Next,0,sizeof(Next));
    int sum = 0;
    for(int i=1;i<=n;i++)
    {
        memset(used,0,sizeof(used));
        if(Find(i))
            sum++;
    }
    return sum;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int k;
    while(cin>>n>>k)
    {
        m=n;
        MEM(line,0);
        for(int i=1;i<=k;i++)
        {
            int x,y;
            cin>>x>>y;
            line[x][y]=1;
        }
        cout<<maxMatch()<<endl;
    }
    return 0;
}