Moving Tables

Problem 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. 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.

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

題意描述：

有400個房間，如圖1排列，一直走廊只能放下一張桌子，一張桌子從一個房間移動到另一個房間需要10分鐘。當移動桌子不共用一個走廊時可以同時進行移動，來減少用時，當走廊衝突時，要分開來移動。給出n個移動，求出最少用時。

解題思路：

由圖1可知，(1,2)，（3,4），（5,6）……每一組為一個走廊，將房間號加1除2，即為走廊號，根據房間號，求出需要使用的走廊號。走廊使用一次對應次數加一，最後求出走廊的最大使用次數乘10即為所求的最少用時。

錯誤分析：

剛看到這道題一直想到的是用陣列，找最後房間號排序，當走廊衝突時再相加時間，WA了好多次，看了題解發現思路錯了。這道題還要注意下先輸入的房間號的大小不一定。

程式程式碼：

#include<stdio.h>
#include<string.h>
int main()
{
	int t,num,n,i,s,x,y,q[250],j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		memset(q,0,sizeof(q));
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&x,&y);
			x=(x+1)/2;
			y=(y+1)/2;
			if(x>y)
			{
				s=x;
				x=y;
				y=s;
			}
			for(j=x;j<=y;j++)
				q[j]++;
		}
		s=0;
		for(i=1;i<=210;i++)
		{
			if(q[i]>s)
				s=q[i];
		}
		printf("%d\n",s*10);
	}
	return 0;
}