POJ 1017 Packets （贪心）

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解决一个工厂产品装箱的问题，通过算法最小化所需包裹数量，达到节省成本的目的。输入为不同大小的产品数量，输出是最少的包裹数。采用贪心算法实现，确保每个大包裹尽可能被充分利用。

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Packets

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 45720		Accepted: 15454

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

经典贪心题目处理贪心的关键是一定能装就装能装的话尝试可以再装

AC代码如下：

//
//  POJ 1017 Packets
//
//  Created by TaoSama on 2015-02-22
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define CLR(x,y) memset(x, y, sizeof(x))

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

//5*5的还能装36-25=11个1*1
//4*4的还能装36-16=20/(2*2)=5个2*2
//3*3情况如b数组
int a[10], b[4] = {0, 5, 3, 1};
int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	ios_base::sync_with_stdio(0);

	while(cin >> a[1]) {
		for(int i = 2; i <= 6; ++i) cin >> a[i];
		if(count(a + 1, a + 7, 0) == 6) break;

		int ans = a[4] + a[5] + a[6] + (a[3] + 3) / 4;
		int n2 = 5 * a[4] + b[a[3] % 4];
		//2*2装满需要36/4 = 9
		if(a[2] > n2)	ans += (a[2] - n2 + 8) / 9;
		int n1 = 36 * (ans - a[6]) - 25 * a[5] - 16 * a[4] - 9 * a[3] - 4 * a[2];
		if(a[1] > n1)	ans += (a[1] - n1 + 35)/36;
		cout << ans << endl;
	}
	return 0;
}