[VJ][貪心]Moving Tables
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：
一個公司有200個房間，奇數在走廊的一側，偶數在走廊的一側，其中奇數和偶數房間共享一個走廊。（1和2 共用 3和4 共用）
現在給出n對房間，從a房間移動桌子到b房間，其中佔用了（a-b）之間的走廊。（3房間移動到5房間間時 3房間 5房間 4房間 6房間 都不可用）
移動一次桌子花費10min，求移動所有桌子後花費的總時間。
正確解法：
剛開始以為是活動分配問題，排序開始的房間和後面的房間，可是奇數房間和偶數房間共用，就很難。
看書發現一種很簡單的解法，桶排序吧。
其中 （a+1）/2 就可以把奇數和偶數房間合併成一個，現在就有了1-100個房間。
每次把 （a+1）/2 到 （b+1）/2 中間的房間全部 +1
最後數其中一個房間用的次數最多的 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 }View Code
給了我一個很新奇的思路，把奇數偶數房間合併之後，就完全可以用任務分配問題了。
等我努力開創出來。
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：
一個公司有200個房間，奇數在走廊的一側，偶數在走廊的一側，其中奇數和偶數房間共享一個走廊。（1和2 共用 3和4 共用）
現在給出n對房間，從a房間移動桌子到b房間，其中佔用了（a-b）之間的走廊。（3房間移動到5房間間時 3房間 5房間 4房間 6房間 都不可用）
移動一次桌子花費10min，求移動所有桌子後花費的總時間。
正確解法：
剛開始以為是活動分配問題，排序開始的房間和後面的房間，可是奇數房間和偶數房間共用，就很難。
看書發現一種很簡單的解法，桶排序吧。
其中 （a+1）/2 就可以把奇數和偶數房間合併成一個，現在就有了1-100個房間。
每次把 （a+1）/2 到 （b+1）/2 中間的房間全部 +1
最後數其中一個房間用的次數最多的 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 }View Code
給了我一個很新奇的思路，把奇數偶數房間合併之後，就完全可以用任務分配問題了。
等我努力開創出來。