Moving Tables

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.

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

10
20
30

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

int main()
{
    int T,N;
    int c[401];
    while (scanf("%d", &T)!=EOF)
    {
        while (T--)
        {
            scanf("%d", &N);        
            int a,b;
            int max = -1;
            memset(c, 0, sizeof(c));
            for(int i=0; i<N; i++)
            {
                scanf("%d %d", &a, &b);
                //将区间限制为从小到大地移动桌子 
                if (a > b)
                {
                    swap(a, b);
                }
                //因为走廊两边的排列都是奇数与偶数相对，而不论什么一个要移动，
                //都会影响对面的不能被使用。也就是shared
                //所以。假设是偶数開始，那么会影响到对面的，也就是a-1
                //同理，假设是奇数结束，也会影响到对面的，也就是b+1
                //那么在这个区间都是被shared的 
                if (a%2 == 0)
                    a--;
                if (b%2 == 1)
                    b++;
                for (int i=a; i<=b; i++)
                {
                    //标记被shared的区间[a, b]，被使用一次加一次，所以是c[i]++ 
                    c[i]++;
                    //找重叠数最大之中的一个，即必须经过的次数 
                    if (c[i]>max)
                        max = c[i];
                }
            }               
            printf("%d\n", max*10);
        }                
    }    
    return 0;
}