B - Moving Tables

最新推荐文章于 2024-07-26 09:15:00 发布

转载最新推荐文章于 2024-07-26 09:15:00 发布 · 162 阅读

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原文链接：http://www.cnblogs.com/NYNU-ACM/p/4248803.html

本文探讨了一个经典的算法问题“MovingTables”，涉及在限定走廊空间中高效移动桌子的策略。通过分析不同移动方案，提出了一种利用计数数组和最大元素搜索的方法来解决此问题，旨在最小化所有桌子移动所需的时间。

B - Moving Tables

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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Description

The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure.

The floor has 200 rooms each on the north side and south side along the corridor. Recently the Company made a plan to reform its system. The reform includes moving a lot of tables between rooms. Because the corridor is narrow and all the tables are big, only one table can pass through the corridor. Some plan is needed to make the moving efficient. The manager figured out the following plan: Moving a table from a room to another room can be done within 10 minutes. When moving a table from room i to room j, the part of the corridor between the front of room i and the front of room j is used. So, during each 10 minutes, several moving between two rooms not sharing the same part of the corridor will be done simultaneously. To make it clear the manager illustrated the possible cases and impossible cases of simultaneous moving.

For each room, at most one table will be either moved in or moved out. Now, the manager seeks out a method to minimize the time to move all the tables. Your job is to write a program to solve the manager’s problem.

Input

The input consists of T test cases. The number of test cases ) (T is given in the first line of the input. Each test case begins with a line containing an integer N , 1<=N<=200 , that represents the number of tables to move. Each of the following N lines contains two positive integers s and t, representing that a table is to move from room number s to room number t (each room number appears at most once in the N lines). From the N+3-rd line, the remaining test cases are listed in the same manner as above.

Output

The output should contain the minimum time in minutes to complete the moving, one per line.

Sample Input

       3 4 10 20 30 40 50 60 70 80 2 1 3 2 200 3 10 100 20 80 30 50 
     

Sample Output

       10 20 30
     

      一直WA，查不出来，，，，后来偶然看到我的for循环是i《begin；i！=end；i++。。。。哎，平常写的手顺了，应该是〈=。下回要特别注意。附上25行代码：
     

      my answer:
     

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int T,n;
    cin>>T;
    while(T--)
    {
	    scanf("%d",&n);
        int m=0,begin,end, sum[500]={0};
        for(int i=0;i!=n;i++){
            cin>>begin>>end;
            begin = (begin - 1)/2;
            end = (end - 1 )/2;
            if(begin >end )swap(begin ,end );
            if(m<end)m=end;
			for(int j=begin;j<=end;j++)
                sum[j]+=10;
        }
        printf("%d\n",*max_element(sum,sum+m));
    }
    return 0;
}

有一点就是房间是要分奇偶的，上面那个有点不理解为啥要减一除以2.下面是我的代码：

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int T,n;
    cin>>T;
    while(T--)
    {
	    scanf("%d",&n);
        int m=0,begin,end, sum[500]={0};
        for(int i=0;i!=n;i++){
            cin>>begin>>end;
			if(begin >end )swap(begin ,end );
            if(end%2!=0) end++;
            if(m<end)m=end;
			for(int j=begin;j<=end;j++)
                sum[j]+=10;
        }
        printf("%d\n",*max_element(sum,sum+m));
    }
    return 0;
}

转载于:https://www.cnblogs.com/NYNU-ACM/p/4248803.html