acm_problem_a_moving_tables

Problem A： moving tables

Description：

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

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想法1：

要进行多次移动桌子，就因为有重叠的走廊被占用，那么走廊有200小段，计算其占用最大次数，即最小移动次数。

输入数据后，将房间号转化为走廊对应次序，用语言整型除法自动略去小数的特点，（n-1）/2；
数据遍历输入，得c[i]占用次数，遍历出其最大值；

#include <iostream>#include <cstring>using namespace std;int main(){    short T,N,s,t,c[200],i;    cin>>T;    while (T--)    {        memset(c,0,sizeof(c));        cin>>N;        while (N--)        {            cin>>s>>t;            s = (s-1)>>1; //(n-1)/2 room->corridor            t = (t-1)>>1;            s < t ? : s^=t^=s^=t;// exchange s&t if (s>t)            i = s;            while(i<=t)            {                ++c[i];                ++i;            }        }        short j = 200;        while(j--)        {            c[j]<c[j-1]? : c[j-1] = c[j];        }        cout<<c[0]*10<<endl;    }    return 0;}

想法2：

贪心：
未用数学证明，
因为是要尽可能少次数移动，贪新标准为使得走廊空余空间最少，可按房间号开始前面空的走廊越少越好,依照struct table里面的s，排序，循环计数，直至不再有桌子排列，输出。
然而代码未通过，通过再贴出来。