poj1083 思维题

http://poj.org/problem?id=1083

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

10
20
30

/**
poj 1083 思维
题目大意；如图所示在一条走廊的两侧各有200个房间，现在给定一些成对的房间相互交换桌子，但是走廊每次只能通过一组搬运，
          也就是说如果两个搬运过程有交叉是不能同时搬运的，要依次来，一次搬运10min，问完成所有的搬运的最少用时。
解题思路：考虑每个房间有多少搬运过程需要经过，我们截取最大的房间经过的次数就可以了，挺锻炼思维的一道题。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

int n,a[405];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(x>y)swap(x,y);
            if(x%2==0)x--;
            if(y%2==1)y++;
            for(int i=x;i<=y;i++)
                 a[i]++;
        }
        int maxx=-1;
        for(int i=1;i<=400;i++)
        {
            maxx=max(maxx,a[i]);
        }
        printf("%d\n",maxx*10);
    }
    return 0;
}