【南理oj】220 - 推桌子（思路，好题）

推桌子

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 3

描述

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.

输入

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.

输出

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

样例输入

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

样例输出

10
20
30

上传者

苗栋栋

刚开始的思路是用贪心的方法找可以同时进行的方案，代码繁琐，时间太长，还容易出错。

后来用了一种巧妙的方法，就是把每次占用的走廊位置都标记，重叠的就重复标记，最后找到重叠次数最多的即是结果。

代码如下：

#include <cstdio>
int main()
{
	int u;
	scanf("%d",&u);
	while(u--)
	{
		int n;
		int st,end;
		int t;
		int p[411]={0};
		scanf("%d",&n);
		int max=0;
		while(n--)
		{
			scanf("%d %d",&st,&end);
			if(st>end)
			{
				t=st;
				st=end;
				end=t;
			}
			if (st&1)
				st=st/2+1;
			else
				st/=2;
			if (end&1)
				end=end/2+1;
			else
				end/=2;
			for(int i=st;i<=end;i++)
			{
				p[i]+=10;
				if(p[i]>max) 
					max=p[i];
			}
		}
		printf("%d\n",max);
	}
	return 0;
}

贪心的思路一直WA，就不发了，这个就很好。