uva 10626 Buying Coke （DP记忆化搜索）

uva 10626 Buying Coke

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

3

2 2 1 1

2 1 4 1

20 200 3 0

Output for Sample

5

3

148

题目大意：有三种硬币1元n1个，5元n5个，10元n10个，要拿去卖n瓶8元的可乐，问如何投币会使得投入的硬币数量最小。

解题思路：每次找回的硬币数量也会是最小的。可以用记忆化搜索做 ,vis[n1][n5][n10]标记当前所剩硬币情况是否出现过，若出现过直接返回dp[n1][n5][n10]

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#define MAX 1<<20
using namespace std;
typedef long long ll;
int vis[700][200][100], dp[700][200][100];
int num, N1, N5, N10;
int DP(int n, int n1, int n5, int n10) {
    if (vis[n1][n5][n10]) {
        return dp[n1][n5][n10];
    } else if (n == 0) {
        vis[n1][n5][n10] = 1;
        dp[n1][n5][n10] = 0;
        return dp[n1][n5][n10];
    } else {
        dp[n1][n5][n10] = MAX;
        if (n1 >= 8) dp[n1][n5][n10] = min(dp[n1][n5][n10], DP(n - 1, n1 - 8, n5,  n10) + 8);
        if (n5 >= 1 && n1 >= 3) dp[n1][n5][n10] = min(dp[n1][n5][n10], DP(n - 1, n1 - 3, n5 - 1, n10) + 4);
        if (n5 >= 2) dp[n1][n5][n10] = min(dp[n1][n5][n10], DP(n - 1, n1 + 2, n5 - 2, n10) + 2);
        if (n10 >= 1) dp[n1][n5][n10] = min(dp[n1][n5][n10], DP(n - 1, n1 + 2, n5, n10 - 1) + 1);
        if (n10 >= 1 && n1 >= 3) dp[n1][n5][n10] = min(dp[n1][n5][n10], DP(n - 1, n1 - 3, n5 + 1, n10 - 1) + 4);
        vis[n1][n5][n10] = 1;
        return dp[n1][n5][n10];
    }
}

int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d %d %d %d", &num, &N1, &N5, &N10);
        memset(vis, 0, sizeof(vis));
        printf("%d\n", DP(num, N1, N5, N10));
    }
    return 0;
}