【贪心算法】POJ-1017

一、题目

Description

A factory produces products packed in square packets of the same height h and of the sizes 11, 22, 33, 44, 55, 66. 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 11 to the biggest size 66. 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

二、思路&心得

• 题目大意为共有1 * 1、2 * 2...6 * 6的产品各a[i]个，包装袋大小固定为6 * 6，问最少最要多少个包装袋，可以把每个订单中所有产品包装起来。
• 这个题目有点类似硬币问题，在选择时从最大的6 * 6的产品开始依次往小进行计算。对于6 * 6、5 * 5、4 * 4的产品，各需要包装袋a[6]、a[5]、a[4]个，其中放置5 * 5产品的包装袋可以额外装11个1 * 1的产品，放置4 * 4产品的可以额外装5个2 * 2的产品；对于3 * 3的产品比较特殊，对4取模后，根据余数0、1、2、3分四种情况进行判断，每次选择时尽可能得装入更多的2 * 2的产品再装入1 * 1的产品；对于2 * 2和1 * 1的产品判断较为简单，不再多说。
• 在做这题时刚开始有点看不懂题意，便看了下discuss区，发现所有人都说这题非常非常难以及细节很多，导致不敢轻易下手。但是思路想清，WA了一两次之后，便AC了，好像也没想象中的那么难。
• 在网上观摩到大神的极短代码，算法思想非常精辟，特另附上，以供学习。

三、代码

我的渣解法：

#include<cstdio>

int a[7];

int ans;

void solve() {
	ans += (a[4] + a[5] + a[6]);
	a[1] -= a[5] * 11;
	a[2] -= a[4] * 5;
	if (a[2] < 0) {
		a[1] += a[2] * 4;
	}
	ans += (a[3] / 4 + 1);
	switch (a[3] % 4) {
		case 0: {
			ans --;
			break;
		}
		case 1: {
			if (a[2] > 0) {
				a[2] -= 5;
				if (a[2] < 0) {
					a[1] += a[2] * 4;
				}
				a[1] -= 7;
			} else {
				a[1] -= 27;
			}
			break;
		} 
		case 2: {
			if (a[2] > 0) {
				a[2] -= 3;
				if (a[2] < 0) {
					a[1] += a[2] * 4;
				}
				a[1] -= 6;
			} else {
				a[1] -= 18;
			}
			break;
		}
		case 3: {
			if (a[2] > 0) {
				a[2] -= 1;
				a[1] -= 5;
			} else {
				a[1] -= 9;
			}
			break;
		}
	}
	if (a[2] > 0) {
		ans += a[2] / 9;
		if (a[2] % 9 > 0) {
			ans ++;
			a[1] -= (9 - a[2] % 9) * 4;			
		}
	}
	if (a[1] > 0) {
		ans += (a[1] + 35) /36;
	}
	printf("%d\n", ans);
}

int main () {
	while (1) {
		ans = 0;
		for (int i = 1; i <= 6; i ++) {
			scanf("%d", &a[i]);
		}
		if (!a[1] && !a[2] && !a[3] && !a[4] && !a[5] && !a[6]) break;
		solve();
	}
	return 0;
}

大神的精辟解法：

#include<stdio.h>
int main()
{
	int n,a,b,c,d,e,f,x,y;
	int u[4]={0,5,3,1};
	while(1)
	{
		scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f);
		if(a==0&&b==0&&c==0&&d==0&&e==0&&f==0)
			break;
		n=d+e+f+(c+3)/4;
		y=5*d+u[c%4];//在已有n个的情况下，能装下y个2*2的
		if(b>y)
			n+=(b-y+8)/9;//把多的2*2的弄进来
		x=36*n-36*f-25*e-16*d-9*c-4*b;
		if(a>x)
			n+=(a-x+35)/36;//把1*1的弄进来
		printf("%d\n",n);
	}
	return 0;
}