Moving Tables

本文介绍了一种解决大型办公环境中高效移动桌子问题的算法。该算法通过分析走廊使用情况,确保多个桌子移动过程中的最小冲突,从而实现整体移动效率的最大化。

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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.

 Table movingReason
Possible( room 30 to room 50) and (room 60 to room 90)no part of corridor is shared
(room 11 to room 12) and (room 14 to room 13)no part of corridor is shared
Impossible(room 20 to room 40) and (room 31 to room 80)corridor in front of room 31 to room 40 is shared
(room 1 to room 4) and (room 3 to room 6) corridor in front of room 3 is shared
(room 2 to room 8) and (room 7 to room 10) corridor in front of room 7 is shared

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.


输入

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 , 1<= N<=200 , that represents the number of tables to move. Each of the following lines contains two positive integers and t, representing that a table is to move from room number to room number (each room number appears at most once in the lines). From the N+3-rd line, the remaining test cases are listed in the same manner as above.


输出

The output should contain the minimum time in minutes to complete the moving, one per line.


样例输入

3
4
10 20
30 40
50 60
70 80
2
1 3
2 200
3
10 100
20 80
30 50

样例输出

10
20
30

#include <stdio.h>
 
int main(){
    int T, N, i, j, sum, from, to;
     
    scanf("%d", &T);
    while(T--){
        scanf("%d", &N);
        int max = 0;
        int a[400] = {0};
        for(j = 0; j < N; j++){
            scanf("%d%d", &from, &to);
        if(from > to){
            int k = from;
            from = to;
            to = k;
        }
        if(from % 2 == 0)
               from -= 1;
            if(to % 2 == 1)
               to +=1;
        for(i = from; i <= to;i++){
            a[i] += 10;
        }
        }
        for(j = 0; j < 400; j++){
            if(max < a[j])max = a[j];
        }
        printf("%d\n", max);
    }
    return 0;
} 

内容概要:本文探讨了在MATLAB/SimuLink环境中进行三相STATCOM(静态同步补偿器)无功补偿的技术方法及其仿真过程。首先介绍了STATCOM作为无功功率补偿装置的工作原理,即通过调节交流电压的幅值和相位来实现对无功功率的有效管理。接着详细描述了在MATLAB/SimuLink平台下构建三相STATCOM仿真模型的具体步骤,包括创建新模型、添加电源和负载、搭建主电路、加入控制模块以及完成整个电路的连接。然后阐述了如何通过对STATCOM输出电压和电流的精确调控达到无功补偿的目的,并展示了具体的仿真结果分析方法,如读取仿真数据、提取关键参数、绘制无功功率变化曲线等。最后指出,这种技术可以显著提升电力系统的稳定性与电能质量,展望了STATCOM在未来的发展潜力。 适合人群:电气工程专业学生、从事电力系统相关工作的技术人员、希望深入了解无功补偿技术的研究人员。 使用场景及目标:适用于想要掌握MATLAB/SimuLink软件操作技能的人群,特别是那些专注于电力电子领域的从业者;旨在帮助他们学会建立复杂的电力系统仿真模型,以便更好地理解STATCOM的工作机制,进而优化实际项目中的无功补偿方案。 其他说明:文中提供的实例代码可以帮助读者直观地了解如何从零开始构建一个完整的三相STATCOM仿真环境,并通过图形化的方式展示无功补偿的效果,便于进一步的学习与研究。
### 7-2 Moving Tables C语言解决方案 #### 算法思路 为了最小化移动所有桌子所需的时间,采用贪心算法是一个有效的方法。核心思想是在任何时刻尽可能多地利用走廊资源来搬运桌子。具体来说: 对于每一个需要移动的桌子,记录下它所占用的走廊区间的起始位置和结束位置,并统计这些区间重叠的最大次数。因为每次只能有一个桌子通过同一段走廊,所以最大重叠数决定了最少需要多少次5分钟才能完成全部桌子的移动。 #### 实现方法 下面展示了一个完整的C语言程序用于求解此问题[^1]。 ```c #include <stdio.h> int main() { int n, m; scanf("%d", &n); while (n--) { // 处理多组测试数据 int i, j, k, t; int begin, end; int a[200] = {0}; // 记录各个时间段是否有桌椅经过 scanf("%d", &m); for (i = 0; i < m; ++i) { scanf("%d%d", &begin, &end); if (begin > end) { // 如果起点大于终点,则交换两者的位置 t = begin; begin = end; end = t; } // 统计每个时间片内的最大并发量 for (k = (begin - 1) / 2; k <= (end - 1) / 2; ++k){ a[k]++; } } int max = -1; // 找到最大的并发数量作为最终的结果乘以单位时间得到总耗时 for (i = 0; i < 200; ++i) { if (a[i] > max) { max = a[i]; } } printf("%d\n", max * 10); // 输出结果并换行 } return 0; } ``` 这段代码实现了上述提到的逻辑流程,能够有效地处理多个案例的数据输入,并给出相应的最优解。需要注意的是,在实际编程过程中应当仔细考虑边界条件以及变量初始化等问题,确保程序运行稳定可靠[^4]。
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