HDUOJ1050 Moving Tables

Moving Tables

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7097Accepted Submission(s): 2396

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
看到一个恐怖的代码
main(t,i,n,b,e) {scanf("%d",&t);while(t--) {int c[200]= {0},m=0;scanf("%d",&n);while(n--) {scanf("%d%d",&b,&e);if(b>e) {i=b;b=e;e=i;}for(i=(b-1)/2;i<=(e-1)/2;++i)++c[i];}for(i=0;i<200;++i)if(m<c[i])m=c[i];printf("%d\n",10*m);}}
以下是正常一点的代码，也很恐怖
#include<iostream>                                                                            
11
|010| #include<stdio.h>                                                                             
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|011| #include<stdlib.h>                                                                            
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|012| #include<string.h>                                                                            
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|013| /*----------------------*Global Variable*----------------------*/                             
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|014|                                                                                               
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|015| int T,N,s,t,times[200];                                                                       
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|016| //*=======================*Main Program*=======================*//                            
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|017| using namespace std;                                                                          
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|018|                                                                                               
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|019| int main(){                                                                                   
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|020| .       scanf("%d",&T);                                                                       
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|021| .       while(T--){                                                                           
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|022| .       .       scanf("%d",&N);                                                               
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|023| .       .       memset(times,0,sizeof(times));                                                
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|024| .       .       while(N--){                                                                   
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|025| .       .       .       scanf("%d%d",&s,&t);                                                  
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|026| .       .       .       s=(s-1)/2;                                                            
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|027| .       .       .       t=(t-1)/2;                                                            
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|028| .       .       .       if(s>t).       .       //当s>t时交换s,t的值                     
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|029| .       .       .         s^=t^=s^=t;                                                         
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|030| .       .       .       for(int i=s;i<=t;++i)                                                 
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|031| .       .       .         times[i]++;                                                         
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|032| .       .       }                                                                             
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|033| .       .       int res=-1;                                                                   
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|034| .       .       for(int i=0;i<200;++i)                                                        
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|035| .       .         res=res<times[i]?times[i]:res;                                              
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|036| .       .       printf("%d\n",res*10);                                                        
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|037| .       }                                                                                     
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|038| }  
【解题思路】这道题最少花多少时间，实际上我们只要考虑哪一段重合度最高，重合度最高的地方，也就是我们至少要移动的次数了。
因为有400间房间，1-2对应一段走廊，3-4对应一段走廊，如此我们可以把走廊分成200段，标记为a[1]-a[200]，之后我们根据输
进的房间序号，就可以算出要用到哪几段的走廊，之后给对应的a[n]值加1就好，最后求出a[n]最大值就是移动的次数了。