798 - 背包问题VII - LintCode

描述
假设你身上有 n 元，超市里有多种大米可以选择，每种大米都是袋装的，必须整袋购买，给出每种大米的重量，价格以及数量，求最多能买多少公斤的大米
题目链接：https://www.lintcode.com/problem/798/
方法一：递归

#include "iostream"
#include "vector"
#include "cstring"
using namespace std;
int process(vector<int> &prices, vector<int> &weight, vector<int> &amounts, int index, int n) {
    if(index < 0 || n < 0){
        return 0;
    }
    int w = 0;
    for(int i = 0; i * prices[index] <= n && i <= amounts[index]; i++){
        w = max(i * weight[index] + process(prices, weight, amounts, index - 1, n - i * prices[index]),
                process(prices, weight, amounts, index - 1, n)
                );
    }
    return w;
}
int main(){
    vector<int> prices{3,2};
    vector<int> weight{300,160};
    vector<int> amounts{1,6};
    int n = 8;
    cout << process(prices, weight, amounts, prices.size() - 1, n);
}

方法二：dp

class Solution {
public:
    /**
     * @param n: the money of you
     * @param prices: the price of rice[i]
     * @param weight: the weight of rice[i]
     * @param amounts: the amount of rice[i]
     * @return: the maximum weight
     */
    int backPackVII(int n, vector<int> &prices, vector<int> &weight, vector<int> &amounts) {
        // write your code here
        int dp[prices.size()][n + 1];
    	memset(dp, 0, sizeof(dp));
    	for(int i = 0; i < prices.size(); i++){
        	for(int j = 0; j <= n; j++){
            	for(int k = 0; k * prices[i] <= j  && k <= amounts[i]; k++){
                	dp[i][j] = max(dp[i][j], k * weight[i] + (i - 1 >= 0 ? dp[i - 1][j - k * prices[i]] : 0));
                	dp[i][j] = max(dp[i][j], i - 1 >= 0 ? dp[i - 1][j] : 0);
            	}
        	}
    	}
    	return dp[prices.size() - 1][n];
    }
};