hdu 1050 Moving Tables（贪心）

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

题目来源 http://acm.hdu.edu.cn/showproblem.php?pid=1050

思路

统计最大的重叠数，再乘以10即可。具体做法：把每个房间之间的走廊作为一个统计单位，当所有的办公桌都搬运完成之后，看看这段走廊到底需要占用多少次，然后统计所有的走廊被占用的最大值max，这个值就是要单独安排的搬运次数，乘以10就是总的搬运时间。

代码

#include
#include
#include
using namespace std;
int main()
{
    int T,n,max,s,t,a[207];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));//将数组a初始化为0
        for(int i=0;it)//确保smax)
            {
                max=a[i];
            }
        }
        printf("%d\n",max*10);//总的搬运时间
    }
    return 0;
}