HDU 1050

題目連結

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

Code

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int result[
 
405];
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        memset(result,0,sizeof(result));//每次進行前先把result[]給清零

        int N;
        cin>>N;

        for(int i=1;i<=N;i++)
        {
            int Begin,End;
            scanf("%d %d",&Begin,&End);//輸入移動的起始位置

            if (Begin>End)  swap(Begin,End);//用swap函式把小的放前面,大的放後面
            if (Begin%2==0) Begin--;//注意!!!
            if (End%2==1) End++;//注意!!!

            for(int j=Begin;j<End;j++)
                result[j]+=1;
        }

        int *ans=std::max_element(result,result+403);//輸出最大值

        cout<<(*ans)*10<<endl;
    }

    return 0;
}

其實就是把經過的路徑都做上標記,經過一次就標記一次,但是因為是一條走廊,兩邊都有房間,所以要注意起始位置的處理;比如,從1搬到5,那麼會從1,2,3,4,5,6的房間前經過.如果從2搬到5,那麼也會從1,2,3,4,5,6房間經過.所以程式碼裡面的if (Begin%2==0) Begin--;//注意!!! if (End%2==1) End++;//注意!!!

就是用來處理起始位置的;