hdu 1050 Moving Tables

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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
 
/*解题分析:此题采用时间连,将房间号排成一列,每次有对应号进入时,对应的时间增加10;
                   关键部分:注意房间号,是奇偶对应的(开始将这个忽略了,一直提交不过),故首先对其处理,之后按照先后顺序递加时间;
                                for(i=0;i<n;i++)
                                 {
                                    scanf("%d%d",&a,&b);
                                           a=(a-1)/2;
                                           b=(b-1)/2;
                                    if(a>b) { temp=a; a=b; b=temp;//按大小排一下 }
                                       for(j=a;j<=b;j++)
                                        { p[j]+=10; }
                                }          
                                      
*/
#include<stdio.h>#include<string.h>#include<algorithm>#define M 205 using namespace std;//  struct information//    {//        int beg;//        int end;//    }boy[10010]; int cmp(int a,int b)  {         return a>b;    }  int main()   {       int T;       int n;       int a,b,temp;       int i,j,k;       int p[M];       scanf("%d",&T);        while(T--)         {              memset(p,0,sizeof(p));              scanf("%d",&n);               for(i=0;i<n;i++)                {                    scanf("%d%d",&a,&b);                a=(a-1)/2;                b=(b-1)/2;            if(a>b)            {                temp=a;                a=b;                b=temp;//按大小排一下            }                          for(j=a;j<=b;j++)                        {                            p[j]+=10;                        }                }                sort(p,p+M,cmp);                              printf("%d\n",p[0]);                        }       return 0;   }

                                                                                                                      
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