<OJ_Sicily>Coins完全背包问题_ouyang has 6 kinds of coins. the number of the i-t-优快云博客

本文链接：https://blog.youkuaiyun.com/Vipin_Pei/article/details/51243402

本文介绍了一个完全背包问题的应用案例，通过使用C++实现了一种算法来解决如何在背包容量限制下，根据拥有的不同种类硬币的数量和各自的价值与重量，计算能够携带的最大价值。该算法采用动态规划的方法，并通过二维数组来存储中间结果。

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Description

Ouyang has 6 kinds of coins.

The number of the i-th coin is N[i] (0<=i<6).

Their value and weight are as follewed:

0. $0.01, 3g

1. $0.05, 5g

2. $0.10, 2g

3. $0.25, 6g

4. $0.50, 11g

5. $1, 8g

Ouyang want to run away from home with his coins.

But he is so weak that he can only carray M gram of coins.

Given the number of each coin he has, what is the maximal value of coins he can take?

Input

There are multiple cases.

Each case has one line with 7 integers: M (1<=M<=10000), A[i], (0<=i<6, 0<=A[i]<=100000).

Output

The maximal value of coins he can take.

问题解释：这相当于是一个完全背包问题

输入：每一个case的输入都是1行，每一行包含7个数，第一个数为背包容量，后面6个数为相应的硬币数量

输出：在背包容量允许的情况下所获得的最大价值

#include <iostream>
#include <algorithm>
#include <iomanip>
#include <memory.h>

using namespace std;
double v[6] = {0.01,0.05,0.10,0.25,0.50,1};  //各硬币的价值
int w[6] = {3,5,2,6,11,8};     //各硬币的重量
int A[6];
double totalV[7][10005];       

int main(int argc, const char * argv[]) {
    int M;
    while (cin >> M) {
        if (M <= 0) break;
        memset(totalV, 0, sizeof(totalV));
        for (int i = 0; i < 6; i ++) {
            cin >> A[i];
        }
        for (int i = 5; i >= 0 ; i--) {
            for (int j = 0; j <= M; j++) {
                totalV[i][j] = totalV[i+1][j];
                for (int k = 1; k <= A[i] ; k++) {   //寻找价值最优解
                    if (k * w[i] > j){               //加入第i个硬币后，重量超重放弃该硬币的加入        
                        break;   
                    }
                    else{  //第i个硬币加入后没有超重，则选择加入或者不加入的最优值
                        //totalV[i][j] = max(totalV[i+1][j],totalV[i][j]);
                        totalV[i][j] = max(totalV[i][j],totalV[i+1][j-k * w[i]]+k * v[i]);
                    }
                }
            }
        }
        cout << "$" << setprecision(2) << setiosflags(ios::fixed) << totalV[0][M] << endl;
    }
    return 0;
}

后记：

在这里还是使用一个二维数组来获取最高价值。每一个问题的最优解都是建立在子问题的最优解的基础上的。