HDU1050（Moving Tables:贪心算法）

解题思路：

    这种做法是基于hdu2037的做法上考虑的，找出所有可以同时搬运的桌子，然后就很方便求出最短总时间。

还有一种更简单的做法是直接遍历一遍找出与别的重复次数最多的那片区域，重复次数*10就可以得到结果。

 

Moving Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28033    Accepted Submission(s): 9170

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

Source

Asia 2001, Taejon (South Korea)

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//#define _CRT_SECURE_NO_DEPRECATE
#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>

using namespace std;

int cmp(const int *a, const int *b) {
    if (a[0] == b[0]) return a[1] < b[1];
    else return a[0] < b[0];
}

int main() {
    int n;
    cin >> n;
    while (n--) {
        int t;
        int **room;
        int time = 0;
        int t1 = 0;
        int flag[256] = { 0 };//用于判断此次搬运是否已经进行。
        room = new int *[256];
        cin >> t;
        for (int i = 0; i < t; i++) {
            room[i] = new int[2];
            cin >> room[i][0] >> room[i][1];
            if (room[i][0] > room[i][1]) {
                int temp = room[i][0];
                room[i][0] = room[i][1];
                room[i][1] = temp;
            }
            if (room[i][1] % 2 == 1) room[i][1]++;//此处是这道题的坑点，因为走廊两侧都有房间，所以
            if (room[i][0] % 2 == 0) room[i][0]--;//3和4号房间公用一片走廊。
        }
        sort(room, room + t, cmp);
        for (int i = 0; i < t; i++) {
            if (flag[i] == 0) {
                time += 10;
                int boundary = room[i][1];
                for (int j = i + 1; j < t; j++) { //找出可以与第i次搬运同时进行的次数
                    if (room[j][0] > boundary && flag[j] == 0) {
                        boundary = room[j][1];
                        flag[j] = 1;
                        /*printf("%d %d %d\n", i, time, boundary);纠错语句*/
                    }
                }
            }
        }
        cout << time << endl;
        /*for (int i = 0; i < t; i++) {
            cout << room[i][0] << room[i][1] << endl;
        }*/
    }
    //system("pause");
    return 0;
}