uva 10626 Buying Coke (DP + 记忆化搜索)

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探讨了在有限硬币条件下购买特定数量可乐所需的最少硬币数，通过动态规划算法实现。

Problem D
Buying Coke
Input: Standard Input

Output: Standard Output

Time Limit: 2 Seconds

I often buy Coca-Cola from the vending machine at work. Usually I buy several cokes at once, since my working mates also likes coke. A coke in the vending machine costs 8 Swedish crowns, and the machine accept crowns with the values 1, 5 and 10. As soon as I press the coke button (after having inserted sufficient amount of money), I receive a coke followed by the exchange (if any). The exchange is always given in as few coins as possible (this is uniquely determined by the coin set used). This procedure is repeated until I've bought all the cokes I want. Note that I can pick up the coin exchange and use those coins when buying further cokes.

Now, what is the least number of coins I must insert, given the number of cokes I want to buy and the number of coins I have of each value? Please help me solve this problem while I create some harder problems for you. You may assume that the machine won't run out of coins and that I always have enough coins to buy all the cokes I want.

Input

The first line in the input contains the number of test cases (at most 50). Each case is then given on a line by itself. A test case consists of four integers: C (the number of cokes I want to buy), n₁, n₅, n₁₀ (the number of coins of value 1, 5 and 10, respectively). The input limits are 1 <=C <= 150, 0 <= n₁ <= 500, 0 <= n₅ <= 100 and 0 <= n₁₀ <= 50.

Output

For each test case, output a line containing a single integer: the minimum number of coins needed to insert into the vending machine.

Sample Input Output for Sample Input

2 2 1 1

2 1 4 1

20 200 3 0

题意：每瓶可乐 8 元，你有 n1个1元，n2 个 5 元，n3 个10 元，需要买 C 瓶可乐（每次只能

买一瓶），要求你用钱的张数最小并输出最小值。

思路：　dp[ i ][ j ][ k ] 表示你有 i 个1元，j 个 5 元，k 个10 元时所花的钱的张数最小值。

然后用记忆化搜索特别地清晰。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int inf=99999999;

int dp[750][155][55],n,m,k,num;

int DP(int coun,int n1,int n2,int n3)
{
     if(coun==0)  return 0;
     if(dp[n1][n2][n3]!=-1) return dp[n1][n2][n3];
     int ans=inf;
     if(n1>=8)           ans=min(ans,DP(coun-1,n1-8,n2,n3)+8);
     if(n2>=2)           ans=min(ans,DP(coun-1,n1+2,n2-2,n3)+2);
     if(n3>=1)           ans=min(ans,DP(coun-1,n1+2,n2,n3-1)+1);
     if(n1>=3 && n2>=1)  ans=min(ans,DP(coun-1,n1-3,n2-1,n3)+4);
     if(n1>=3 && n3>=1)  ans=min(ans,DP(coun-1,n1-3,n2+1,n3-1)+4);
     if(ans==inf)  ans=0;
     return dp[n1][n2][n3]=ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d %d",&num,&n,&m,&k);
        memset(dp,-1,sizeof(dp));
        printf("%d\n",DP(num,n,m,k));
    }
    return 0;
}