【原】 POJ 1083 Moving Tables 解题报告

http://poj.org/problem?id=1083

方法：
将所有房间号分为200段，即数组path[201]，将每次需要移动的起始和目的地之间的所有path[i]都加1
找出path[1...N-1]中最大的一个，即最大的某路段重叠数，就是需要的趟数，再乘以10即得结果

注意点：
1.由于奇偶数分列两边，所以需要将奇偶数房间号做一下变换：
    如果值是奇数，则该值除以2再+1；如果是偶数直接除以二，此时号的范围变成1~200
2.可能源号比目的号大

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

#include <iostream>
#include <fstream>

using namespace std ;

const int N = 201 ;


void run1083()
{
    ifstream in("in.txt");

    int n,m ;
    int start,end ;
    int i;
    int count ;
    in>>n ;  //number of the test cases
    while( n--!=0 && in>>m )  //number of pathes needed to be move
    {
        unsigned char path[N]={0} ;  //record the number of each segmentation been passed
        while( m--!=0 && in>>start>>end )
        {
            if(start>end)    //if the start bigger than end , swap them
            {int tmp=start; start=end; end=tmp;}
            start = start%2==0 ? start/2 : start/2+1 ;  //change the index
            end = end%2==0 ? end/2 : end/2+1 ;

            for(i=start ; i<=end ; ++i)  //record every segmentation been passed
                ++path[i] ;
        }
        count = path[1] ;
        for( i=2 ; i<N ; ++i )  //find the largest segmentation
            count = path[i]>count ? path[i] : count ;
        cout<<count*10<<endl;
    }
}