Moving Tables
Description
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.
Examples
Input
3
4
10 20
30 40
50 60
70 80
2
1 3
2 200
3
10 100
20 80
30 50
Output
10
20
30
描述：
一个公司有200个房间，奇数在走廊的一侧，偶数在走廊的一侧，其中奇数和偶数房间共享一个走廊。（1和2 共用 3和4 共用）
现在给出n对房间，从a房间移动桌子到b房间，其中占用了（a-b）之间的走廊。（3房间移动到5房间间时 3房间 5房间 4房间 6房间 都不可用）
移动一次桌子花费10min，求移动所有桌子后花费的总时间。
正确解法：
刚开始以为是活动分配问题，排序开始的房间和后面的房间，可是奇数房间和偶数房间共用，就很难。
看书发现一种很简单的解法，桶排序吧。
其中 （a+1）/2 就可以把奇数和偶数房间合并成一个，现在就有了1-100个房间。
每次把 （a+1）/2 到 （b+1）/2 中间的房间全部 +1
最后数其中一个房间用的次数最多的 就是答案。
1 #include "pch.h" 2 #include<iostream> 3 #include<cstdio> 4 #include<string> 5 #include<cstring> 6 #include<cmath> 7 using namespace std; 8 int main() 9 { 10 int t,n; 11 cin >> t; 12 int a[210]; 13 while (t--) { 14 int n; 15 long long maxx = 0; 16 cin >> n; 17 memset(a, 0, sizeof(a)); 18 for (int i = 1; i <= n; i++) 19 { 20 int b,c; 21 cin >>b>>c; 22 for (int j = (b + 1) / 2; j <= (c + 1) / 2; j++) 23 a[j]++; 24 } 25 for (int i = 1; i <= 100; i++) 26 if (a[i] > maxx) maxx = a[i]; 27 cout << maxx * 10 << endl; 28 } 29 return 0; 30 }
给了我一个很新奇的思路，把奇数偶数房间合并之后，就完全可以用任务分配问题了。
等我努力开创出来。
Description
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.
Examples
Input
3
4
10 20
30 40
50 60
70 80
2
1 3
2 200
3
10 100
20 80
30 50
Output
10
20
30
描述：
一个公司有200个房间，奇数在走廊的一侧，偶数在走廊的一侧，其中奇数和偶数房间共享一个走廊。（1和2 共用 3和4 共用）
现在给出n对房间，从a房间移动桌子到b房间，其中占用了（a-b）之间的走廊。（3房间移动到5房间间时 3房间 5房间 4房间 6房间 都不可用）
移动一次桌子花费10min，求移动所有桌子后花费的总时间。
正确解法：
刚开始以为是活动分配问题，排序开始的房间和后面的房间，可是奇数房间和偶数房间共用，就很难。
看书发现一种很简单的解法，桶排序吧。
其中 （a+1）/2 就可以把奇数和偶数房间合并成一个，现在就有了1-100个房间。
每次把 （a+1）/2 到 （b+1）/2 中间的房间全部 +1
最后数其中一个房间用的次数最多的 就是答案。
1 #include "pch.h" 2 #include<iostream> 3 #include<cstdio> 4 #include<string> 5 #include<cstring> 6 #include<cmath> 7 using namespace std; 8 int main() 9 { 10 int t,n; 11 cin >> t; 12 int a[210]; 13 while (t--) { 14 int n; 15 long long maxx = 0; 16 cin >> n; 17 memset(a, 0, sizeof(a)); 18 for (int i = 1; i <= n; i++) 19 { 20 int b,c; 21 cin >>b>>c; 22 for (int j = (b + 1) / 2; j <= (c + 1) / 2; j++) 23 a[j]++; 24 } 25 for (int i = 1; i <= 100; i++) 26 if (a[i] > maxx) maxx = a[i]; 27 cout << maxx * 10 << endl; 28 } 29 return 0; 30 }
给了我一个很新奇的思路，把奇数偶数房间合并之后，就完全可以用任务分配问题了。
等我努力开创出来。