【贪心】【HDOJ-1050】Moving Tables

最新推荐文章于 2020-01-26 17:01:26 发布

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最新推荐文章于 2020-01-26 17:01:26 发布

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原文链接：http://www.cnblogs.com/ahu-shu/p/3551829.html

本文介绍了一种解决办公室间高效搬动桌子的问题算法。通过计算最大交点次数来确定最少搬桌时间，确保走廊内同时只有一张桌子移动。算法采用C++实现，考虑了各种边界情况。

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

/**************************************************************************************************************
题意:各个办公室之间要搬桌子，每次搬动需要10分钟，并且在某张桌子搬动期间，那段走道只能有这一张桌子在搬动
例如要从2-5 那么 1-6 1-4 3-5都不可以搬
思路：求出搬动的过程中最大的交点的次数即可
注意：1、可能a > b 即从16搬到10
2、4-5搬动期间 3和6都是不可用的
**************************************************************************************************************/
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
int map[200+10];
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(map , 0 , sizeof(map));
        int a, b, ans = 0;
        int n;
        scanf("%d", &n);
        while(n--)
        {
            scanf("%d %d", &a, &b);
            if(a > b)
                swap(a , b);
            for(int i = (a + 1) / 2; i <= (b + 1) / 2; i++)
                map[i]++;
        }
        for(int i = 0 ; i < 201 ; i ++)
            ans = max(ans, map[i]);
        printf("%d\n", ans * 10);
    }
    return 0;
}
 