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
Output for the Sample Input
10
20
30
题意及分析：
在一层楼房里有编号1-400的房间，奇数号和偶数号分列两边。现在有一些桌子要移动，但是走廊的原因，一段路只能容纳下一张桌子的移动工作。（也就是说不同路段就可以同时进行）有一些桌子要从一个房间移动到另一些房间，要你求所需的最短的时间。
贪心的想法是：先按房间的编号排序，把没有冲突的移动找出来。把有冲突的移动保存。然后再从保存的移动里找出那些互相不矛盾的移动。直到所有移动全部完成。
AC代码：
还有一种思路就是，把处在对门的两个编号变为一个编号。那么这些1-400的编号就变成了1-200的编号。再然后计算每一个房间所处路段经过的次数，取其中的最大值。
代码如下：
hdu 1050 Moving Tables,布布扣,bubuko.com
原文：http://blog.csdn.net/is_cp/article/details/38705539