zoukankan      html  css  js  c++  java
  • poj 3041(最大匹配问题)

    http://poj.org/problem?id=3041
    Asteroids
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 14476   Accepted: 7880

    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.
     

    题目大意是导弹能够消灭直线上的障碍物,给出障碍物的位置,问最少须要多少导弹,把矩阵中的行放一边。列放还有一边,按二分图做,就不是非常难了,这一类的题目主要就是主函数那里有点不一样,其它套着模板来。

    模板题

    AC代码:

    #include <stdio.h>
    #include <string.h>
    #define MAXN 510
    
    int N,K,ans;
    bool map[MAXN][MAXN];
    int match[MAXN];
    bool vis[MAXN];
    
    bool find(int u)
    {
    	int i;
    	for ( i=1; i<=N; i++)
    	{
    		if (map[u][i] && !vis[i] )
    		{
    			vis[i] = true;
    			if ( match[i] == -1 || find(match[i]) )
    			{
    				match[i] = u;
    				return true;
    			}
    		}
    	}
    	return false;
    }
    
    int main()
    {
    	int i,x,y;
    	ans = 0;
    	scanf("%d%d",&N,&K);
    	memset(map,0,sizeof(map));
    	memset(match,-1,sizeof(match));
    	for (i=0; i<K; i++)
    	{
    		scanf("%d%d",&x,&y);
    		map[x][y] = 1;
    	}
    	for (i=1; i<=N; i++)
    	{
    		memset(vis,0,sizeof(vis));
    		if (find(i))
    			ans++;
    	}
    	printf("%d
    ",ans);
    	return 0;
    }



  • 相关阅读:
    教你玩转CSS的所有字体,路过的来瞧一瞧。
    CSS基本语法?
    怎么样才能玩转前端所有的CSS背景相关问题?
    如何玩转CSS内部样式表与外部样式表?
    什么是CSS?你真的理解了吗?
    CSS页面demo
    HTML教程(看完这篇就够了)
    HTML5新增的主体结构元素
    Math
    练习点菜及用的公式
  • 原文地址:https://www.cnblogs.com/lcchuguo/p/5401797.html
Copyright © 2011-2022 走看看