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:
    域名申请攻略(以godaddy+支付宝为例)
    初始java白盒测试junit的使用
    微型oracle学习使用—oracle XE(oracle express edition
    VBS学习创建桌面快捷方式
    强烈推荐Oracle的入门心得
    Godaddy域名注册详细图文教程(转)
    如何用WordPress搭建自己的博客(转)
    图解eclipse+myeclipse完全绿色版制作过程
    java整理的经典的bug问题白盒问题(转)
    eclipse插件整理集合(包括myeclipse插件)转
  • 原文地址:https://www.cnblogs.com/zbtrs/p/8127116.html
Copyright © 2011-2022 走看看