Question

以下是this SPOJ问题的尝试解决方案。输入是：

硬币中一定金额的总重量
使用货币的硬币的价值和相应的权重，目标是找到给定金额的最小可能货币价值。

我稍微修改了wikipedia article关于背包问题的背包问题的动态编程解决方案 - 我刚刚按重量对硬币进行了分类，所以我不必通过所有硬币来获取最小值和（希望）确保组合重量等于容量。（请看代码，它非常简单并且评论过。）

然而，根据判断，有一个输入，我的算法给出了错误的答案。你能否说一下这个算法有什么问题？

#include <iostream>
#include <vector>
#include <algorithm>
#include <limits>

using namespace std;

typedef unsigned int weight_t;
typedef unsigned int value_t;

struct coin {
    weight_t weight;
    value_t value;
};

coin make_coin(weight_t weight, value_t value) {
    coin ret;
    ret.weight = weight;
    ret.value = value;
    return ret;
}

bool compare_by_weight(const coin& coin1, const coin& coin2) {
    return coin1.weight < coin2.weight;
}

int main() {
    unsigned int test_cases;
    cin >> test_cases;
    while(test_cases--) {
        //Initialization
        unsigned int number_of_coins, n = 0;
        weight_t empty_pig, full_pig, coin_weight, coins_weight;
        value_t coin_value, min_value, current_value = 0;
        vector<coin> coins;
        vector<unsigned int> min_value_for_the_weight;

        //Input
        cin >> empty_pig >> full_pig;
        cin >> number_of_coins;
        n = number_of_coins;
        while(n--) {
            cin >> coin_value >> coin_weight;
            coins.push_back(make_coin(coin_weight, coin_value));
        }

        //Input processing
        coins_weight = full_pig - empty_pig;
        sort(coins.begin(), coins.end(), compare_by_weight);
        min_value_for_the_weight.resize(coins_weight+1);
        for(unsigned int i = 0; i < coins_weight; i++) min_value_for_the_weight[i] = 0;

        //For all weights
        for(unsigned int i = 1; i <= coins_weight; i++) {
            //Find the smallest value
            min_value = numeric_limits<value_t>::max();
            for(unsigned int j = 0; j < number_of_coins; j++) {
                //The examined coin weights more or same than examined total weight and we either already have put a coin
                //in, or this is the first one
                if(coins[j].weight <= i && (min_value_for_the_weight[i - coins[j].weight] > 0 || i == coins[j].weight)){
                    current_value = coins[j].value + min_value_for_the_weight[i - coins[j].weight];
                    if(current_value < min_value) min_value = current_value;
                } else break;  // <- this I deleted to get accepted
            }
            if(min_value == numeric_limits<value_t>::max()) min_value = 0;
            min_value_for_the_weight[i] = min_value;
        }

        //If the piggy empty, output zero
        if(!min_value_for_the_weight[coins_weight] && coins_weight != 0)
            cout << "This is impossible." << endl;
        else
            cout << "The minimum amount of money in the piggy-bank is " << min_value_for_the_weight[coins_weight] << "." << endl;
        }
    return 0;
}

Answer 1

案例empty_pig == full_pig是有问题的，因为你忽略了重复min_value_for_the_weight的第0条的逻辑特殊大小。

另一个问题是，如果break coins[j].weight > i只是个好主意。当break和另一半合并为假时，旧代码可coin[j].weight <= i，即无法制作重量为i - coins[j].weight的硬币。这发生在以下测试用例中。

我们必须使用重量为2或3的硬币来制作重量3.重量1被正确地确定为不可能。正确地确定重量2是可能的。对于重量3，我们尝试重量为2的硬币，确定重量1是不可能的，并且在尝试重量-3硬币之前break。结果是代码错误地报告说不可能做出重量3。

改进的动态背包 - 有问题的输入？

1 个答案: