# poj 3041 Asteroids 最小点覆盖/最大匹配

## Asteroids

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16242 Accepted: 8833

## 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.

3 4
1 1
1 3
2 2
3 2

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).

## 代码

``#define REP(i, n) for (int i=0;i<n;++i)#define RD(n) scanf("%d",&n)int V;vector<int> G[maxn];int match[maxn];int used[maxn];void add_edge(int u,int v){    G[u].push_back(v);    G[v].push_back(u);}bool dfs(int v){    used[v]=1;    for(int i=0;i<G[v].size();i++)    {        int u=G[v][i],w=match[u];        if(w<0||!used[w]&&dfs(w))        {            match[v]=u;            match[u]=v;            return 1;        }    }    return 0;}int bipartite_matching(){    int res=0;    memset(match,-1,sizeof(match));    for(int v=0;v<V;v++)    {        if(match[v]<0)        {            memset(used,0,sizeof(used));            if(dfs(v))                res++;        }    }    return res;}int n,k;int R[maxn],C[maxn];void solve(){    V=n*2;    for(int i=0;i<k;i++)    {        add_edge(R[i]-1,n+C[i]-1);    }    printf("%d\n",bipartite_matching());}int main(){    scanf("%d%d",&n,&k);    REP(i,k)    {        RD(R[i]);        RD(C[i]);    }    solve();    return 0;}`` 