HDU ACM 1050 Moving Tables

原题描述

Moving Tables

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

解题思路

转为求最大重叠次数。使用一个数组p[200]作为走廊每一格被经过的次数，找到最大的乘10即是答案

参考代码

#include <iostream>#include <algorithm>using namespace std;int main(){    int T, N, p[200], s, e;    cin >> T;    while(T--)    {        cin >> N;        memset(p, 0, sizeof(p));        for (int i = 0; i < N;i++)        {            cin >> s >> e;            s = (s - 1) / 2;            e = (e - 1) / 2;                // 将坐标转化为统一的走廊坐标            if (s > e)                      // 如果输入的前一个数比后一个大            {                int temp = s;                s = e;                e = temp;            }            for (int j = s; j <= e; j++)    // 走廊坐标需要经过一次，该坐标加一            {                p[j]++;            }        }        int max = -1;                       // 找出经过次数最多的点(的最大重叠次数)        for (int i = 0; i < 200;i++)        {            if (max < p[i])                max = p[i];        }        cout << max * 10 << endl;    }}