Buying Coke - UVa 10626 dp

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

3 2 2 1 1 2 1 4 1 20 200 3 0

5 3 148

题意：一瓶可乐8元钱，每次只能买一瓶可乐，然后你手上分别有1元，5元，10元硬币若干枚，每买完一瓶后，他会用最少的硬币数给你找钱，问你最少向里面投多少枚硬币。

思路：注意有一种情况，就是你投3个1元的和1个10元的，他会找你一个5元的，如果没有这种情况的话就可以直接贪心了，但是这道题还是只能dp。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[710][210][110],vis[710][210][110],T,t;
void dfs(int n,int num1,int num2,int num3)
{ if(vis[num1][num2][num3]==t)
   return;
  vis[num1][num2][num3]=t;
  if(n==0)
  { dp[num1][num2][num3]=0;
    return;
  }
  dp[num1][num2][num3]=1000000000;
  if(num1>=8)
  { dfs(n-1,num1-8,num2,num3);
    dp[num1][num2][num3]=min(dp[num1][num2][num3],dp[num1-8][num2][num3]+8);
  }
  if(num1>=3 && num2>=1)
  { dfs(n-1,num1-3,num2-1,num3);
    dp[num1][num2][num3]=min(dp[num1][num2][num3],dp[num1-3][num2-1][num3]+4);
  }
  if(num2>=2)
  { dfs(n-1,num1+2,num2-2,num3);
    dp[num1][num2][num3]=min(dp[num1][num2][num3],dp[num1+2][num2-2][num3]+2);
  }
  if(num1>=3 && num3>=1)
  { dfs(n-1,num1-3,num2+1,num3-1);
    dp[num1][num2][num3]=min(dp[num1][num2][num3],dp[num1-3][num2+1][num3-1]+4);
  }
  if(num3>=1)
  { dfs(n-1,num1+2,num2,num3-1);
    dp[num1][num2][num3]=min(dp[num1][num2][num3],dp[num1+2][num2][num3-1]+1);
  }
}
int main()
{ int i,j,k,ans,ret,p,n,num1,num2,num3;
  scanf("%d",&T);
  for(t=1;t<=T;t++)
  { scanf("%d%d%d%d",&n,&num1,&num2,&num3);
    dfs(n,num1,num2,num3);
    printf("%d\n",dp[num1][num2][num3]);
  }
}