动态规划(二)HDU1114

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发布于 2019-02-21 17:18:01

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发布于 2019-02-21 17:18:01

1.题目来源HDU1114

Sample Input 3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4

Sample Output The minimum amount of money in the piggy-bank is 60. The minimum amount of money in the piggy-bank is 100. This is impossible.

2.题目分析

题目首先给定一个空存钱罐的重量和这个存钱罐最多能装进去的重量，现在需要在不打破这个存钱罐的情况下猜测里面最少的钱。每种钱的数量不做限制，条件是必须装满，同时给出每种钱币的价值和重量。参考背包九讲，这属于完全背包问题，初始状态为装满的情况，所以放钱币的策略就变成了放几个问题,有的钱币组合不合理就设置得比较大。

3.代码如下

#include<iostream>
#include<algorithm>
using namespace std;
const int num = 10005;
const int abc = 9999999999;
int main(void)
{
    //(1 <= P <= 50000, 1 <= W <= 10000)
    //1 <= E <= F <= 10000, N (1 <= N <= 500)(coins)
    int dp[num], C[num], W[num], T, N, E, F, V, i, j;
    cin >> T;
    while (T-- > 0) {
        cin >> E >> F;   // the weight of an empty pig and of the pig filled with coins
        cin >> N;       //the number of various coins used in the given currency. 
        V = F - E;      //the rest of the space
                        //C is the value of the coin in monetary units, W is it's weight in grams. 
        for (i = 1; i <= N; i++)
            cin >> W[i] >> C[i];    
        dp[0] = 0;
        for (i = 1; i <= V; i++) dp[i] = abc;
        for (i = 1; i <= N; i++) {
            for (j = C[i]; j <= V; j++) {
                dp[j] = min(dp[j], dp[j - C[i]] + W[i]);
            }
        }
        if (dp[V] == abc) cout << "This is impossible."<<endl;
        else cout << "The minimum amount of money in the piggy-bank is "<<dp[V]<<"."<<endl;
    }
    return 0;
}

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原始发表：2018年03月10日，如有侵权请联系 cloudcommunity@tencent.com 删除

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