poj-1083-Moving Tables-思维-java

Moving Tables

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 33654		Accepted: 11245

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

Source

Taejon 2001

   解题思路：

                                    题意这样的 分奇偶房间 每次给出起点和终点房间号 在运东西经过房间的过程中 不能同时再运别的东西

                                    就是多条线交叉 每次只能走一条线

                                    问 把所有的东西运完最少的时间花费是多少

                                    解法就是对路径上的房间经过次数进行计数 找出最大的那个数

                                     然后乘10 就是花费的最小时间

                                     为什么呢？ 因为 经过次数最多 能走完 那么其他的东西肯定也能运完

                                     

import java.util.Scanner;

public class Main{

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);
		int t = scanner.nextInt();
		for (int i = 0; i < t; i++) {
			int n = scanner.nextInt();
			int [] arr = new int [420];
			int sum = 0;
			for (int j = 0; j < n; j++) {
				int a = scanner.nextInt();
				int b = scanner.nextInt();
				if(a > b) {
					a = a + b;
					b = a - b;
					a = a - b;
				}
				if (a % 2 == 0) {
					a --;
				}
				if (b % 2 != 0) {
					b ++;
				}
				for (int k = a; k <= b; k++) {
					arr[k] ++;
					sum = Math.max(sum, arr[k]);
				}
			}
			System.out.println(sum*10);
		}
	}

}