poj 1017 Packets 模拟题

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本文介绍了一种针对特定尺寸产品的最优包装算法，旨在通过合理安排不同尺寸的产品来减少包装盒的使用数量，从而降低运输成本。该算法特别关注了如何有效地利用空间，特别是针对较大尺寸产品的包装策略。

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Packets

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41736		Accepted: 14018

Description

A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square parcels of the same height h as the products have and of the size 6*6. Because of the expenses it is the interest of the factory as well as of the customer to minimize the number of parcels necessary to deliver the ordered products from the factory to the customer. A good program solving the problem of finding the minimal number of parcels necessary to deliver the given products according to an order would save a lot of money. You are asked to make such a program.

Input

The input file consists of several lines specifying orders. Each line specifies one order. Orders are described by six integers separated by one space representing successively the number of packets of individual size from the smallest size 1*1 to the biggest size 6*6. The end of the input file is indicated by the line containing six zeros.

Output

The output file contains one line for each line in the input file. This line contains the minimal number of parcels into which the order from the corresponding line of the input file can be packed. There is no line in the output file corresponding to the last ``null'' line of the input file.

Sample Input

0 0 4 0 0 1 
7 5 1 0 0 0 
0 0 0 0 0 0

Sample Output

2 
1

Source

Central Europe 1996

本题大意为把6种不同尺寸（分别是1*1，2*2，3*3，4*4，5*5，6*6）不同数量的产品用6*6的包装盒来包装，求最小包装盒数。主要关键在于弄清题意，然后先包装尺寸大的，其他剩余空间用尺寸小的产品补充，尤其注意讨论3*3尺寸包装的剩余空间补足情况。

#include <cstdio>

//本题主要考察模拟包装物品操作
int square[7]; //表示目前包装盒剩余空间对应的size数量
int parcel[7]; //表示输入的产品尺寸对应的数量

int main()
{
    bool flag = true;
    int num = 0;
    while(1)
    {
        flag = true;
        for(int i = 1; i <= 6 ; i++)
        {
            scanf("%d", &parcel[i]);
            if(parcel[i] != 0)
                flag = false;
        }
        if(flag)
            break;
        num = parcel[6] + parcel[5] + parcel[4]; //包装尺寸为6*6和5*5和4*4的只能增加包装盒的数量
        num = num + parcel[3] / 4;
        square[3] = 0;
        if(parcel[3] % 4 != 0) //假如尺寸为3*3的包装没有装满包装盒
        {
            num ++;
            square[3] = 4 - (parcel[3] % 4);
        }
        square[1] = 11 * parcel[5]; //包装5*5产品所剩下的1*1尺寸个数，可供1*1产品使用
        square[2] = 5 * parcel[4];//包装4*4产品所剩2*2尺寸个数,可供2*2和1*1使用
        if(square[3] == 1 || square[3] == 0)
        {
             square[2] = square[2] + square[3];//剩下的能容纳2*2的空间
             square[1] = square[1] + square[3] * 5;//目前只能容纳1*1的空间数量
        }
        else if(square[3] == 2)
        {
            square[2] = square[2] + 3;
            square[1] = square[1] + 6;

        }
        else
        {
            square[2] = square[2] + 5;
            square[1] = square[1] + 7;
        }
        if(parcel[2] <= square[2])
            square[1] = square[1] + 4 * (square[2] - parcel[2]);
        else
        {
                parcel[2] = parcel[2] - square[2];
                num = num + parcel[2] / 9;
                if(parcel[2] % 9 != 0)
                {
                    num++;
                    square[1] += 4 * (9 - (parcel[2] % 9));
                }
         }
        if(square[1] < parcel[1])
        {
            parcel[1] = parcel[1] - square[1];
            num = num + parcel[1] / 36;
            if(parcel[1] % 36 != 0)
                num = num + 1;
        }
        printf("%d\n", num);
    }
	return 0;
}