POJ Moving Tables

Time Limit: 1000MS Memory Limit: 10000K 

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 file. 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 3 + N -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

102030

Source

Taejon 2001

题意：

一家公司有400间办公室，需要将给定房间的桌子进行移动。约定，过道只能同时移动一张桌子，每次移动需要耗费10s钟。问在给定的n组数据中需要花费多长的时间。

我的解法：

自我感觉这题还是比较简单的。过道只能允许一张桌子移动，于是如果有两次移动经过相同的“路径”，那就不能同时移动。基于此，只需将“每段路径”上可能移动的桌子次数加以统计，问题就迎刃而解了。

400个房间分成两排，可将整个路径细分成200段。统计没分段上移动的桌子次数。

我的代码：

#include <iostream>
#include <fstream>

using namespace std;

int state[201];

int main()
{
 int T,N;
 int i,j;

// ifstream cin("test.txt");
 cin>>T;

 while( T-- )
 {
  cin>>N;
  int from,to;

  memset( state,0,sizeof(state) );

  int max=0;
  for( i=0; i<N; i++ )
  {
   cin>>from>>to;

   if( from>to )    // 如果from大于to  交换from和to
   {
    from += to;
    to = from-to;
    from = from-to;
   }

   from = (from+1)/2;
   to = (to+1)/2;

   for( j=from; j<=to; ++j )
   {
    state[j]++;
    if( state[j]>max )
     max=state[j];
   }
  }

  cout<<10*max<<endl;
 }
 return 0;
}