poj 1083 Moving Tables

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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 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.
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

题目大意

一层里面有400个房间，北边和南边各有200个房间，要从一个房间里面把一张桌子移动到另一个房间，需要占用这两个房间之间的所有走廊（包括这两个房间前面的），每移动一个桌子需要10分钟，给出需要移动的桌子的数据（从哪移动到哪），要求计算出最少需要多少分钟才能把所有桌子移动完。

不知为何这题是DP。。。用的比较贪的思想，如果从m移动到n（假设m<n，但是在程序中处理输入时需要判断两个数大小），则把序号为(m-1)/2到（n-1）/2的所有数组元素都+10，这样处理完每个桌子后，遍历整个数组寻找最大的一个元素，即为实际的需要时间。。。

#include
#include
using namespace std;
int a[300];
int T,maxs,n,x,y;
void swap(int &a,int &b){int t=a;a=b,b=t;}
main()
{
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof a);
        scanf("%d",&n);
        maxs=0;
        for (int i=1;i<=n;i++) { scanf("%d%d",&x,&y); if (x>y) swap(x,y);
            for (int j=(x-1)/2;j<=(y-1)/2;j++)
                a[j]+=10;
        }
        for(int i=0;i<200;i++)
            if (maxs<a[i]) maxs=a[i];
        printf("%dn",maxs);
    }
}