UVA - 311 Packets（贪心）

智能算法优化的包裹打包问题解决方案

最新推荐文章于 2018-11-10 23:01:00 发布

原创最新推荐文章于 2018-11-10 23:01:00 发布 · 1k 阅读

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本文探讨了如何通过智能算法解决工厂生产并配送不同尺寸的包裹至客户时，如何最小化所需的配送箱数量的问题。通过分析各种可能的打包方案，文章提供了优化算法的实现细节，包括对不同尺寸包裹的合理分配，以及对特定尺寸包裹的特殊处理策略。这种方法不仅节省了成本，也提高了物流效率。

Packets

A factory produces products packed in square packets of the same height h and of the sizes , , , , , . These products are always delivered to customers in the square parcels of the same height h as the products have and of the size . 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 to the biggest size . 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：
一个工厂生产包裹，有6种包裹，分别为1*1、2*2、3*3……6*6，每种包裹的高度相同都为h，现在要将这些包裹装进6*6的箱子中，问最少需要多少个6*6的箱子。

解析：贪心，不难想到应该先装大的包裹，再装小的包裹。
所以按照这个思路来分析一下该题。不妨把每种情况画出来。

(1)6*6，1个6*6
(2)5*5，1个5*5+11个1*1
(3)4*4，1个4*4+5个2*2
(4)3*3应该分4种情况讨论，

4个3*3

1个3*3+5个2*2+7个1*1

2个3*3+3个2*2+6个1*1

3个3*3+1个2*2+5个1*1

(5)2*2
9个2*2
(6)1*1
36个1*1

注意上面的每种情况，每次用到2*2的包裹就要判断，如果2*2的包裹用完的话，要用1*1的包裹来填充。

#include <cstdio>
#include <cstring>
using namespace std;
int a[7];
void judge() {
	if(a[2] < 0) {
		a[1] += 4 * a[2];
		a[2] = 0;
	}
}
int main() {
	int sum,tmp;
	while(1) {
		bool ok = false;
		sum = 0;
		for(int i = 1; i <= 6; i++) {
			scanf("%d",&a[i]);
			ok = ok || a[i];
		}
		if(!ok) {
			break;
		}
		//6*6
		sum += a[6];
		//5*5
		sum += a[5];
		a[1] -= 11 * a[5];
		//4*4
		sum += a[4];
		a[2] -= 5 * a[4];
		judge();
		//3*3
		sum += a[3] / 4;
		tmp = a[3] % 4;
		if(tmp) {
			sum++;
			switch(tmp) {
				case 1:
					a[2] -= 5;
					a[1] -= 7;
					judge();
					break;
				case 2:
					a[2] -= 3;
					a[1] -= 6;
					judge();
					break;
				case 3:
					a[2] -= 1;
					a[1] -= 5;
					judge();
					break;
			}
		}
		//2*2
		if(a[2] > 0) {
			sum += a[2] / 9;
			tmp = a[2] % 9;
			if(tmp) {
				sum++;
			}
			a[1] -= (36 - tmp*4);
		}
		//1*1
		if(a[1] > 0) {
			sum += a[1] / 36;
			tmp = a[1] % 36;
			if(tmp) {
				sum++;
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}