zoukankan      html  css  js  c++  java
  • hdoj 4883 TIANKENG’s restaurant【贪心区间覆盖】

    TIANKENG’s restaurant

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 1360    Accepted Submission(s): 545


    Problem Description
    TIANKENG manages a restaurant after graduating from ZCMU, and tens of thousands of customers come to have meal because of its delicious dishes. Today n groups of customers come to enjoy their meal, and there are Xi persons in the ith group in sum. Assuming that each customer can own only one chair. Now we know the arriving time STi and departure time EDi of each group. Could you help TIANKENG calculate the minimum chairs he needs to prepare so that every customer can take a seat when arriving the restaurant?
     
    Input
    The first line contains a positive integer T(T<=100), standing for T test cases in all.

    Each cases has a positive integer n(1<=n<=10000), which means n groups of customer. Then following n lines, each line there is a positive integer Xi(1<=Xi<=100), referring to the sum of the number of the ith group people, and the arriving time STi and departure time Edi(the time format is hh:mm, 0<=hh<24, 0<=mm<60), Given that the arriving time must be earlier than the departure time.

    Pay attention that when a group of people arrive at the restaurant as soon as a group of people leaves from the restaurant, then the arriving group can be arranged to take their seats if the seats are enough.
     
    Output
    For each test case, output the minimum number of chair that TIANKENG needs to prepare.
     
    Sample Input
    2
    2
    6 08:00 09:00
    5 08:59 09:59
    2
    6 08:00 09:00
    5 09:00 10:00
     
    Sample Output
    11
    6
    题并不难,
    题意:一组数据中每个开头是一个整数代表此组人的人数,接下来是此组人的到达时间和走的时间,如果两组人的
            时间区间不相交,则第二组人可以使用第一组人用过的桌子,如果相交,就必须重新安排桌子,问最少需要多少桌子
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define MAX 10010
    using namespace std;
    struct node
    {
    	int come;
    	int go;
    	int peo;
    }s[MAX];
    int dp[MAX];
    bool cmp(int a,int b)
    {
    	return a>b;
    }
    int main()
    {
    	int t,n,m,j,i;
    	int ha,hb,ma,mb;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%d",&n);
    		memset(dp,0,sizeof(dp));
    		for(i=0;i<n;i++)
    		{
    			scanf("%d %d:%d %d:%d",&s[i].peo,&ha,&ma,&hb,&mb);
    		    s[i].come=ha*60+ma;
    		    s[i].go=hb*60+mb;
    		    for(j=s[i].come;j<s[i].go;j++)
    		    {
    		    	dp[j]+=s[i].peo;
    		    }
    		}
    		sort(dp,dp+MAX,cmp);
    		printf("%d
    ",dp[0]);
    	}
    	return 0;
    }
    

      

  • 相关阅读:
    虚方法、重写方法以及抽象类的知识小结
    DateTime时间格式
    JavaScript中Eval()函数的作用
    JQuery Event属性说明
    正则表达式30分钟入门教程
    dwz的form表单中url的变量替换
    dwz中权限的控制
    Dwz下拉菜单的二级联动
    Win7下用IIS发布网站
    IntelliJ IDEA 常用快捷键列表及技巧大全
  • 原文地址:https://www.cnblogs.com/tonghao/p/4691170.html
Copyright © 2011-2022 走看看