贪心算法--Packets

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针对工厂需要将不同尺寸的产品高效打包成统一规格的包裹问题，本文介绍了一种算法，通过计算不同尺寸产品的数量来确定所需的最小包裹箱数，旨在减少运输成本。

POJ:http://poj.org/problem?id=1017

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

题目大意:所有的箱子都是6*6尺寸的，但是物品有1*1，2*2，3*3，4*4，5*5以及6*6尺寸的，要求把所有的物品放进箱子中，求最少需要多少箱子。
题目思路:对于6*6，5*5以及4*4尺寸的物品每个物品需要占有一个箱子，对于3*3的物品一个箱子可以放4个，2*2的物品箱子可以放9个，1*1的可以放36个。采用面积统计1*1箱子的空位，采用向上去整的方法统计箱子。先放置6*6，5*5、4*4、3*3向上取整得到所需箱子n1，统计剩余空格子（4*4、3*3尺寸的物品所在箱子），然后2*2尺寸的物品放入装有4*4、3*3的箱子(这一步要注意3*3所在的箱子剩余的能容纳2*2物品的数量)，空间不够准备新箱子n2（向上取整），统计（n1+n2）剩余空格子最后放入1x1尺寸的箱子，不够空间增加新箱子n3(向上取整)。总需箱子n1+n2+n3.
#include<iostream>
#include<cstdio>
#include<cstring>
/*程序思路:
先算出3X3 4X4 5X5 6X6需要n的箱子;
1X1 2X2填充其中;
1X1 2X2不过填充则另外启用新箱子m.
总的箱子=m+n;

(value + (n-1)) / n;程序技巧:向上取整
*/
int num[4] = { 0, 5, 3, 1 };//放入3X3
int box[7];
int default_input[7] = { 0, 0,0,4,0,0,1};
//int default_input[7] = { 0, 7,5,1,0,0,0};

int main(){

	//freopen("input.txt","r",stdin);

	while (1){
		int tmp = 0;
		for (int i = 1; i <= 6; i++){
			//scanf("%d", &box[i]);    // 读入每个物品的数目
			box[i] = default_input[i];
			tmp += box[i];
		}
		if (tmp == 0)
			break;
		int ans = box[6] + box[5] + box[4] + (box[3] + 3) / 4;  //a6,a5,a4,每个物品占有一个箱子(a3 + 3 ) / 4 代表a3的物品需要占
		int a2 = box[4] * 5 + num[box[3] % 4];  //统计3X3 4X4 5X5 6X6放进箱子中后a2物品的空位子有多少；
		if (box[2]>a2)
			ans += (box[2] - a2 + 8) / 9;
		int a1 = ans * 36 - box[6] * 36 - box[5] * 25 - box[4] * 16 - box[3] * 9 - box[2] * 4;//统计2X2 3X3 4X4 5X5 6X6放入后容纳a1物品的空位；
		if (box[1]>a1)   //求1x1的空位子，只需要统计剩余的面积即可；
			ans += (box[1] - a1 + 35) / 36;
		printf("%d\n", ans);
	}
	return 0;
}