poj 1276 [多重背包]

最新推荐文章于 2021-02-26 20:09:32 发布

Vectorxj

最新推荐文章于 2021-02-26 20:09:32 发布

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分类专栏：动态规划文章标签： poj

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本文介绍了一个关于CashMachine的算法问题，该问题是关于如何通过不同面额的纸币组合来达到最接近请求金额的现金数额。文章详细解释了输入输出格式，并提供了一段C++代码示例，展示了如何通过多重背包问题的思路解决这一问题。

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Cash Machine
Description
A Bank plans to install a machine for cash withdrawal. The machine is able to deliver appropriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example, N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10 means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each. Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine. Notes: @ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.
Input
The program input is from standard input. Each data set in the input stands for a particular transaction and has the format: cash N n1 D1 n2 D2 … nN DN where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct.
Output
For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.
Sample Input
735 3 4 125 6 5 3 350
633 4 500 30 6 100 1 5 0 1
735 0
0 3 10 100 10 50 10 10

大意就是有一堆物品，每种物品有相应的个数，每个物品有相应的花费，还有总钱数。
问消费最大但不超过总钱数的钱数是多少。
多重背包：
if (f[j]) f[j + k * a[i].cost] = 1
那么见代码：

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cstring>
#define M 100050
#define N 15
using namespace std;

int m, n, k, sum;
bool f[M];
struct object {
    int cost, num;
    inline void read(void) {
        scanf("%d%d", &num, &cost);
    }
}a[N];

int main(void) {
    while (~scanf("%d", &m)) {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) a[i].read();
        if (!(n | m)) {
            puts("0"); continue;
        }
        memset(f, 0, sizeof(f));
        f[0] = 1;
        for (int i = 1; i <= n; i++)
            for (int j = m; j >= 0; j--) 
                if (f[j]) {//如果!f[j]就不需要转移了
                    for (k = 1; (sum = a[i].cost * k + j) <= m && k <= a[i].num; k++) {
                        f[sum] = 1;
                    }
                }
        for (k = m; !f[k]; k--);
        printf("%d\n", k);
    }
    return 0;
}