HDU - 1050 Moving Tables (greedy algorithm)

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

Problem solving ideas:

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Two rooms with opposite doors occupy the same corridor, so for example, when there are 1-3 and 4-6 movement modes, they share the same corridor, that is, the corridor in front of the door of room 3.

The analysis process is shown in the following figure:

The above analysis process is transferred from : (transferred from https://blog.csdn.net/code_pang/article/details/8251240 )

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After looking at the picture analysis, it is easy to come up with a solution to the problem.
1. Divide the corridor into 200 small sections according to the rooms, for example, rooms 1 and 2 correspond to the first section of corridor, and rooms 3 and 4 correspond to the second section of corridor... .
2. A map array is defined in the code of this question, which is used to record the number of times each corridor is occupied, and find the corridor that is occupied the most times. The final time required is the maximum number of times this corridor is occupied * 10.

Code:

#include <cstdio>
#include <cstring>
using namespace std;


typedef struct{
    int s;
    int t;
}Movement;

int minvalue(int a , int b){
    if(a < b)
        return a;
    return b;
}

int maxvalue(int a , int b){
    if(a > b)
        return a;
    return b;
}

Movement mov[210];

int map[210];
int main(){
    freopen("D://testData//1050.txt" , "r" , stdin);
    int t , n , i , j , max;
    scanf("%d",&t);
    while(t --){
        memset(map , 0 , sizeof(map));
        scanf("%d",&n);
        max = 0;
        for(i = 0 ; i < n ; i ++){
            scanf("%d %d",&mov[i].s , &mov[i].t);
        }

        for(i = 0 ; i < n ; i ++){

            //计算这次移动会占据哪一段路 ， 因为肯能是从左向右移动 ， 也有可能从右向左移动 ，这里统一看成左-->右
            int temp1 = (minvalue(mov[i].s , mov[i].t) + 1 )/ 2;
            int temp2 = (maxvalue(mov[i].s , mov[i].t) + 1 )/ 2;

            for(j = temp1 ; j <= temp2 ; j ++){
                map[j] = map[j] ++;
                if(map[j] > max)
                    max = map[j];
            }
        }

        printf("%d\n",10*max);
    }
}