zoukankan      html  css  js  c++  java
  • 【二分图匹配入门专题1】G

    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
    题意:有一个n*n的矩阵,输入k行数,每行数是一个点的坐标,一次性可以去掉矩阵一行或者一列的~~那个单词没查,反正就是去掉这些坐标点,问你最少的操作次数
    思路:就是一个最小顶点覆盖的模板题,和杭电的2119一样。
    #include<stdio.h>
    #include<string.h>
    #define N 550
    
    int n,k;
    int book[N],e[N][N],match[N];
    
    int dfs(int u)
    {
        int i;
        for(i = 1; i <= n; i ++)
        {
            if(!book[i]&&e[u][i])
            {
                book[i] = 1;
                if(!match[i]||dfs(match[i]))
                {
                    match[i] = u;
                    return 1;
                }
            }
        }
        return 0;
    }
    
    int main()
    {
        int t1,t2,sum,i,j;
        while(scanf("%d%d",&n,&k)!=EOF)
        {
            sum = 0;
            memset(e,0,sizeof(e));
            memset(match,0,sizeof(match));
            for(i = 1; i <= k; i ++)
            {
                scanf("%d%d",&t1,&t2);
                e[t1][t2] = 1;
            }
            for(i = 1; i <= n; i ++)
            {
                memset(book,0,sizeof(book));
                if(dfs(i))
                    sum ++;
            }
            printf("%d
    ",sum);
        }
        return 0;
    }
  • 相关阅读:
    internet连接共享被启用时 出现了一个错误 (null)
    mybatis01-1测试
    配置没有问题,虚拟机Ubuntu系统ifconfig没有网卡信息
    Ubuntu启动Apache
    VM虚拟机Linux系统eth0下面没有inet和inet6
    jQuery通过id和name获取值的区别
    1.4.3 ID遍历爬虫(每天一更)
    mysql中的SQL语句执行的顺序
    Mecanim动画系统丶
    html中常见的行内元素和块级元素,还有常见的行内块元素
  • 原文地址:https://www.cnblogs.com/hellocheng/p/7353212.html
Copyright © 2011-2022 走看看