HDU 3591 The trouble of Xiaoqian(多重背包+完全背包）

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本文介绍了一道关于使用最少数量的硬币完成购物支付的问题。主人公小倩希望通过精确计算使用最少的硬币来购买所需商品。面对不同面额的硬币及有限的携带数量，如何在不超过一定数额的情况下完成支付成为挑战。文章提供了使用多重背包算法解决此问题的代码实现。

象盐城大数据竞赛！

The trouble of Xiaoqian

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2451 Accepted Submission(s): 867

Problem Description

In the country of ALPC , Xiaoqian is a very famous mathematician. She is immersed in calculate, and she want to use the minimum number of coins in every shopping. (The numbers of the shopping include the coins she gave the store and the store backed to her.)
And now , Xiaoqian wants to buy T (1 ≤ T ≤ 10,000) cents of supplies. The currency system has N (1 ≤ N ≤ 100) different coins, with values V1, V2, ..., VN (1 ≤ Vi ≤ 120). Xiaoqian is carrying C1 coins of value V1, C2 coins of value V2, ...., and CN coins of value VN (0 ≤ Ci ≤ 10,000). The shopkeeper has an unlimited supply of all the coins, and always makes change in the most efficient manner .But Xiaoqian is a low-pitched girl , she wouldn’t like giving out more than 20000 once.

Input

There are several test cases in the input.
Line 1: Two space-separated integers: N and T.
Line 2: N space-separated integers, respectively V1, V2, ..., VN coins (V1, ...VN)
Line 3: N space-separated integers, respectively C1, C2, ..., CN
The end of the input is a double 0.

Output

Output one line for each test case like this ”Case X: Y” : X presents the Xth test case and Y presents the minimum number of coins . If it is impossible to pay and receive exact change, output -1.

Sample Input

Sample Output

      
       Case 1: 3

应用多重背包求解dp[T]-dp[20000],即xioaqian付T~20000分别需要的硬币数，然后用完全背包求解dp2[1]-dp2[20000],

dp2[j]为店主找j元钱所需要的硬币数，因此只需找出最小的dp[j]+dp2[j-T]即可。

注意：初始化时将dp设为无穷大，付0元需要0个硬币，所以dp[0]=0,dp2[0]=0;这样最后若无法通过付钱找钱得到T,则dp[j]+dp2[j-T]为无穷大。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
int dp[20003],INF=1e9,dp2[20002];
using namespace std;
int main()
{
    int N, T, ca = 1;
    while (scanf_s("%d%d", &N, &T) != EOF)
    {
        if (!(N + T))break;
        int v[102], c[102];
        for (int i = 1; i <= N; i++)
            scanf_s("%d", &v[i]);
        for (int i = 1; i <= N; i++)
            scanf_s("%d", &c[i]);
        for (int i = 0; i <= 20000; i++)
            dp[i] = INF, dp2[i] = INF;
        dp[0] = 0;
        for (int i = 1; i <= N; i++)
        {
            int k = 1, M = c[i];
            while (k < M)
            {
                for (int j = 20000; j >=k*v[i]; j--)
                {
                    dp[j] = min(dp[j], dp[j - k * v[i]] + k);
                }
                M = M - k; k = 2 * k;
            }
            for (int j = 20000; j >= M * v[i]; j--)
            {
                dp[j] = min(dp[j], dp[j - M * v[i]] + M);
            }
        }
        dp2[0] = 0;
        for (int i = 1; i <= N; i++)
        {
            for (int j = v[i]; j <= 20000; j++)
            {
                dp2[j] = min(dp2[j], dp2[j - v[i]] + 1);
            }
        }
        int mi = INF;
        for (int i = T; i <= 20000; i++)
        {
            int t = i - T;
            if (dp[i] + dp2[t] < mi) mi = dp[i] + dp2[t];
        }
        if (mi == INF) printf("Case %d: -1\n", ca);
        else printf("Case %d: %d\n",ca, mi);
        ca++;
    }
    return 0;
}