POJ1017 Packets(贪心算法训练)
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 51306 Accepted: 17391
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
一、题目大意
公司共有底面面积为1*1、2*2、3*3、4*4、5*5、6*6，高度同为H的六种产品，现在需要用最少的箱子打包，箱子的底面面积为6*6，高度为H。
二、解题思路
简单的暴力贪心算法，对不同的产品有不同的策略，按照从大到小的顺序打包产品，策略如下：
6*6：1个产品放在1个箱子里
5*5：1个产品要占用1个箱子，用1*1的箱子可以填充（11个填满1箱）
4*4：1个产品要占用1个箱子，剩余空间用2*2和1*1的箱子填充（先填充2*2，再填充1*1）
3*3：4个产品可以填满1个箱子，假如有不满1个箱子的，分情况用1*1和2*2的产品填满
2*2：9个产品可以填满1个箱子，假如有不满1个箱子的，用1*1的产品填充
1*1：36个产品可填满一个箱子
三、具体代码
1 #include <cstdio> 2 3 int MAX_(int a, int b){ 4 if(a>b) return a; 5 else return b; 6 } 7 8 int main(){ 9 int s1, s2, s3, s4, s5, s6; 10 while(scanf("%d%d%d%d%d%d", &s1, &s2, &s3, &s4, &s5, &s6) && s1+s2+s3+s4+s5+s6){ 11 int packets = 0; 12 packets += s6; // 6*6的产品一个装一箱 13 14 packets += s5; // 5*5的产品一个装一箱 15 s1 = MAX_(0, s1-11*s5); // 剩余空间用1*1的产品尽量填满 16 17 packets += s4; // 4*4的产品一个装一箱 18 if(s2<5*s4) s1 = MAX_(0, s1-(5*s4-s2)); // 假如2*2的产品填完之后仍然有空隙，则用1*1填满 19 s2 = MAX_(0, s2-5*s4); // 尽量用2*2的产品填满 20 21 packets += (s3+3)/4; // 3*3的产品四个一箱 22 s3 %= 4; // 假如3*3的箱子不是四的倍数个，则先用2*2填充再用1*1填充 23 if(s3==1){ 24 if(s2<5) s1 = MAX_(0, s1-(27-4*s2)); 25 else s1 = MAX_(0, s1-7); 26 s2 = MAX_(0, s2-5); 27 } 28 else if(s3==2){ 29 if(s2<3) s1 = MAX_(0, s1-(18-4*s2)); 30 else s1 = MAX_(0, s1-6); 31 s2 = MAX_(0, s2-3); 32 } 33 else if(s3==3){ 34 if(s2<1) s1 = MAX_(0, s1-(9-4*s2)); 35 else s1 = MAX_(0, s1-5); 36 s2 = MAX_(0, s2-1); 37 } 38 39 packets += (s2+8)/9; // 2*2的产品九个一箱 40 s2 %= 9; // 假如2*2的箱子不是九的倍数个，则用1*1填充 41 if(s2) s1 = MAX_(0, s1-(36-4*s2)); 42 43 packets += (s1+35)/36; // 1*1的产品三十六个一箱 44 45 printf("%d\n", packets); 46 } 47 48 return 0; 49 }
