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 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.
The output should contain the minimum time in minutes to complete the moving, one per line.
1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cstring>
5 using namespace std;
6 int ans,sum[207];
7 int t,n;
8 int main()
9 {
10 //freopen("in.txt","r",stdin);
11 int from,to;
12 scanf("%d",&t);
13 while(t--){
14 scanf("%d",&n);
15 memset(sum,0,sizeof(sum));
16 ans=0;
17 for(int i=0;i<n;i++)
18 {
19 scanf("%d%d",&from,&to);
20 if(from%2==0) from--;
21 if(to%2==0) to--;
22 if(from/2>to/2) swap(from,to);
23 for(int j=from/2;j<=to/2;j++)
24 {
25 sum[j]++;
26 ans=max(ans,sum[j]);
27 }
28 }
29 printf("%d\n",ans*10);
30 }
31 return 0;
32 }
