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.

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.

The output should contain the minimum time in minutes to complete the moving, one per line.

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

题意：
要你搬桌子，有重叠的区域段不能同时搬，不重叠可以同时。每次搬桌子需要10分钟，问你最少用多长时间。
注意：
1.判断首尾，一开始就比较下，小的为start，大的为end。
                if(start>end){/*要判断头尾*/
                    t=end;
                    end=start;
                    start=t;
                }
2.由于（1，2）（3，4）。。。这种属于同一区段，因此简化下，就是( +1)/2,将区域划分，因为偶数加上没影响，所以可以不判断都直接计算。
                start=(start+1)/2;
                end=(end+1)/2;　　　　　
3.核心算法：贪心：哪个区域的经过次数最多，该次数就是我们所求取得数值；
　　　　　　　　　for(j=start;j<=end;j++)
                    a[j]++;
4.排序，排序后得到最大即：最繁忙的区域段 sort(a,a+201,cmp);（由大到小）
5.输出时，不要忘了*10。
代码：
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#include<string.h>
bool cmp(int a,int b)
{
    if (a>b) return true;
    return false;
}
int main(){
    int T,a[201],n,t;
    int i,start,end,j;
    while(scanf("%d",&T)!=EOF){
        while(T--){
            memset(a,0,sizeof(a));/*头文件是<string.h>*/
            scanf("%d",&n);
            for(i=0;i<n;i++){
                scanf("%d%d",&start,&end);
                if(start>end){/*要判断头尾*/
                    t=end;
                    end=start;
                    start=t;
                }
                start=(start+1)/2;
                end=(end+1)/2;
                for(j=start;j<=end;j++)
                    a[j]++;
            }
            sort(a,a+201,cmp);
            printf("%d\n",a[0]*10);
        }
    }
    return 0;
}