poj 3041 Asteroids 最小点覆盖/最大匹配-优快云博客

本文介绍了一个 NxN 网格中利用最少次数的行或列清除操作来消除所有障碍物（小行星）的算法。通过构建二分图并使用最大匹配算法找到最少的操作次数。

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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.

Sample Input

3 4
1 1
1 3
2 2
3 2

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).

题意

给你一个n*n的方阵，然后有一些东西在一些坐标上，你需要用最少的炸弹把这些东西都炸掉，然后问你，最少炸弹数量是多少

题解

我们建立一个二分图，然后把每行当成一个点，把每列当成一个点，如果这一行这一列有炸弹的话，我们就建一条边。那么这道题，就转化成了，最小点覆盖问题，直接跑一大最大匹配就好了
总而言之，言而总之，我也是套版做的= =

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;
}