ACM: 贪心题 poj 1083

Moving Tables

Description

The famous ACM(Advanced Computer Maker) Company has rented a floor of a buildingwhose shape is in the following figure.

The floor has 200 rooms each on the north side and south side alongthe corridor. Recently the Company made a plan to reform itssystem. The reform includes moving a lot of tables between rooms.Because the corridor is narrow and all the tables are big, only onetable can pass through the corridor. Some plan is needed to makethe moving efficient. The manager figured out the following plan:Moving a table from a room to another room can be done within 10minutes. When moving a table from room i to room j, the part of thecorridor between the front of room i and the front of room j isused. So, during each 10 minutes, several moving between two roomsnot sharing the same part of the corridor will be donesimultaneously. To make it clear the manager illustrated thepossible cases and impossible cases of simultaneous moving.

For each room, at most one table will be either moved in or movedout. Now, the manager seeks out a method to minimize the time tomove all the tables. Your job is to write a program to solve themanager's problem.

Input

The input consistsof T test cases. The number of test cases ) (T is given in thefirst line of the input file. Each test case begins with a linecontaining 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 andt, representing that a table is to move from room number s to roomnumber 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 asabove.

Output

The output shouldcontain the minimum time in minutes to complete the moving, one perline.

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: 一个建筑物里面有400个房间, 一半在北边, 另一半在南边, 一一对应, 现在两边之间有一条过道,
      现在移动n个房间里面的桌子, 过道只可以每次通过一张桌子, 从一个房间到另外一个房间耗时10Mins
      现在要求你求出n个移动中最小的时间.

解题思路:
        1. 这题比较特殊, 可以用贪心法做. 一开始再考虑动态规划方向做题.设每个房间为一个节点之类.
        2. 但是仔细想了一下, 只需要考虑每个房间结果的最大重叠数即可.
        问题分析:
                (1). 因为移动中, 只要出现重叠的情况时候就要分开在两个时段中进行.
                (2). 因此需要将每个房间看成一个个点, 统计出重叠的次数.
                (3). 最后就是房间1-2，3-4，5-6...一一对应.
                     当s是偶数时, s--; 当t是奇数时, t++;

代码:
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 405

int n;
int a[MAX];

int main()
{
//    freopen("input.txt","r",stdin);
    int caseNum;
    scanf("%d",&caseNum);
    while(caseNum--)
    {
        scanf("%d",&n);
        int start, end;
        memset(a,0,sizeof(a));
        for(int i = 0; i < n; ++i)
        {
            scanf("%d %d",&start, &end);
            if(start > end)
            {
                int temp = start;
                start = end;
                end= temp;
            }
            
            if(start % 2 == 0) start--;
            if(end % 2 != 0) end++;
            for(int j = start; j <= end; ++j)
                a[j]++;
        }
        int result = 0;
        for(int i = 0; i < MAX; i++)
            result = result > a[i] ? result : a[i];
        printf("%d\n",result*10);
    }
    return 0;
}