Moving tables

最新推荐文章于 2020-11-21 14:32:26 发布

穆梓先生

最新推荐文章于 2020-11-21 14:32:26 发布

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本文介绍了一个关于在限定条件下，如何优化多个桌子在狭窄走廊中移动的问题。通过有效的规划，实现多桌子同时移动以减少总时间。

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Moving Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36594 Accepted Submission(s): 12040

Problem 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

Sample Output

Source

Asia 2001, Taejon (South Korea)

题意：

南北各200间房间，中间有一走廊。现要将若干特定房间的椅子搬到指定房间，一条走廊只能允许一把椅子过去，已知每个椅子从任意房间到任意房间的时间都是10分钟，求经过最优分配后所需的最短时间。

输入：第一行测试个数，第二行要搬几（n）把椅子，后面分别输入椅子的起始房间和终点房间号（n组每组一行）。

输出：一共所需最短时间。

分析：

1号房间和2号房间都对应位置1的走廊，3号房间和4号房间对应位置2的走廊，一共有200个这样的位置。我用a[201]定义这些位置，其中某位置的值就是经历这个位置的次数。如果从某一号房到另一号房，所经历的每个位置的次数都+1，最后再全体扫描下哪个位置经历次数最多，输出值便是这个位置的值*10。不要遗忘可能起始房间号比终止房间号大：

Ex：
1
3
10 20
30 20
100 10
*********
30

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;


int vis[202];


struct Node{
	int x;
	int y;
}node[203];

int main(){
	int t;
	scanf("%d",&t);
	
	while(t--){
		
		int n;
		scanf("%d",&n);		
		memset(vis,0,sizeof(vis));
			
		int a,b;
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&a,&b);
			if(a>b){
				int temp=b;
				b=a;
				a=temp;
			}
			if(a&1)
			a+=1;
			if(b&1)
			b+=1;
			node[i].x=a/2;
			node[i].y=b/2;
			
		}
		for(int j=0;j<n;j++)
		{			
		for(int i=node[j].x;i<=node[j].y;i++)
		vis[i]++;	
		}
		
		
		
		int max=-1;
		for(int i=0;i<201;i++)
		if(vis[i]>max)
		max=vis[i];

		
		printf("%d\n",max*10);		
	}
}