zoukankan      html  css  js  c++  java
  • poj3041 Asteroids

    Asteroids
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 24203   Accepted: 13112

    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

    题目大意:每一次可以将一行或一列的X全部变成O,问最少多少次能够使所有的X变成O.
    分析:二分图匹配挺容易想到的吧.比较容易想到的想法就是对于点(x,y),x向y连一条有向边,接着思路就很容易地出来了.
    #include <cstdio>
    #include <queue>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    int n,k,a[510][1010],flag[1010],pipei[1010],ans;
    
    bool dfs(int u)
    {
        for (int i = n + 1; i <= n * 2; i++)
        {
            if (!flag[i] &&a[u][i])
            {
                flag[i] = 1;
                if (!pipei[i] || dfs(pipei[i]))
                {
                    pipei[i] = u;
                    return true;
                }
            }
        }
        return false;
    }
    
    int main()
    {
        scanf("%d%d",&n,&k);
        for (int i = 1; i <= k; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            a[x][y + n] = 1;
        }
        for (int i = 1; i <= n; i++)
        {
            memset(flag,0,sizeof(flag));
            if(dfs(i))
                ans++;
        }
        printf("%d
    ",ans);
    
        return 0;
    }
  • 相关阅读:
    Centos设置时区和时间同步
    CentOS 安装 Docker
    Docker部署RabbitMq
    Docker部署MinIO对象存储服务
    Docker配置阿里云镜像加速
    IDEA集成Docker部署项目容器
    VMware配置静态ip
    Vue 报错 "TypeError: Cannot read property 'name' of undefined"
    持续交付-发布可靠软件的系统方法
    软件测试总结
  • 原文地址:https://www.cnblogs.com/zbtrs/p/8127116.html
Copyright © 2011-2022 走看看