UVA 311 Packets(贪心 + 模拟)

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 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 题意:有6种包。分别从1X1 到 6X6,每种有一定数量。现在有一些6X6箱子。要求最少需要几个箱子才能装得下。 思路: 贪心。。从最大的开始装。每次都尽量把空隙填满。。模拟这个装的过程即可 #include <stdio.h> #include <string.h> int b[7], num, empty[7]; int solve() { while (b[6]) {//从最大开始装。。。 b[6] --; num ++; } while (b[5]) { b[5] --; if (b[1] >= 11) b[1] -= 11; else b[1] = 0; num ++; } while (b[4]) { b[4] --; if (b[2] >= 5) b[2] -= 5; else { empty[1] = (5 - b[2]) * 4; b[2] = 0; if (b[1] >= empty[1]) b[1] -= empty[1]; else b[1] = 0; } num ++; } memset(empty, 0, sizeof(empty)); while (b[3]) { if (b[3] >= 4) b[3] -= 4; else { if (b[3] == 1) empty[2] = 5; if (b[3] == 2) empty[2] = 3; if (b[3] == 3) empty[2] = 1; empty[1] = 36 - 9 * b[3]; b[3] = 0; if (b[2] >= empty[2]) { b[2] -= empty[2]; empty[1] -= empty[2] * 4; if (b[1] >= empty[1]) b[1] -= empty[1]; else b[1] = 0; } else { empty[1] -= b[2] * 4; b[2] = 0; if (b[1] >= empty[1]) b[1] -= empty[1]; else b[1] = 0; } } num ++; } memset(empty, 0, sizeof(empty)); while (b[2]) { if (b[2] >= 9) b[2] -= 9; else { empty[1] = (9 - b[2]) * 4; b[2] = 0; if (b[1] >= empty[1]) b[1] -= empty[1]; else b[1] = 0; } num ++; } memset(empty, 0, sizeof(empty)); while (b[1]) { if (b[1] >= 36) b[1] -= 36; else b[1] = 0; num ++; } return num; } int main() { while (~scanf("%d%d%d%d%d%d", &b[1], &b[2], &b[3], &b[4], &b[5], &b[6]) && b[6] + b[1] + b[2] + b[3] + b[4] + b[5] != 0) { num = 0; memset(empty, 0, sizeof(empty)); printf("%d\n", solve()); } return 0; }

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