UVA 10626 Buying Coke(dp + 记忆化搜索)

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：给定n瓶可乐，和1元，5元，10元硬币的个数。求出最少投币次数买可乐。

思路：dp，记忆化搜索， 可乐瓶数不用记录，要注意一种特殊情况10块 + 3块可以买一瓶返回5块，开数组要 多开一些用来保存找回的硬币

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int t;
int num, n[3], dp[705][220][100], vis[705][220][100];

int dfs(int num, int n1, int n2, int n3) {
    int &sum = dp[n1][n2][n3];
    int &flag = vis[n1][n2][n3];
    if (flag) 
	return sum;
    else if (num == 0) {
	flag = 1;
	sum = 0;
	return sum;
    }
    else {
	sum = 2000000000;
	if (n1 >= 8) sum = min(sum, dfs(num - 1, n1 - 8, n2, n3) + 8);
	if (n2 >= 1 && n1 >= 3) sum = min(sum, dfs(num - 1, n1 - 3, n2 - 1, n3) + 4);
	if (n3 >= 1 && n1 >= 3) sum = min(sum, dfs(num - 1, n1 - 3, n2 + 1, n3 - 1) + 4);
	if (n2 >= 2) sum = min(sum, dfs(num - 1, n1 + 2, n2 - 2, n3) + 2);
	if (n3 >= 1) sum = min(sum, dfs(num - 1, n1 + 2, n2, n3 - 1) + 1);
	flag = 1;
	return sum;
    }
}

int main() {
    scanf("%d", &t);
    while (t --) {
	memset(vis, 0, sizeof(vis));
	scanf("%d", &num);
	for (int i = 0; i < 3; i ++)
	    scanf("%d", &n[i]);
	printf("%d\n", dfs(num, n[0], n[1], n[2]));
    }
    return 0;
}