uva 10626

dp 记忆化搜索

3个1元和1个10元的情况不能少

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 20010
#define INF 0x7fffffff
#define inf 10000000
#define ull unsigned long long
#define ll long long
using namespace std;
int a[710][110][100];

int dp(int num, int n1, int n2, int n3)
{
    int &ans = a[n1][n2][n3];
    if(ans >= 0) return ans;
    if(num == 0) return ans = 0;
    else
    {
        ans = INF;
        if(n1 >= 3 && n3 >= 1) ans = min(ans, dp(num-1, n1-3, n2+1, n3-1) + 4);
        if(n1 >= 8) ans = min(ans, dp(num-1, n1-8, n2, n3) + 8);
        if(n1 >= 3 && n2 >= 1) ans = min(ans, dp(num-1, n1-3, n2-1, n3) + 4);
        if(n2 >= 2) ans = min(ans, dp(num-1, n1+2, n2-2, n3) + 2);
        if(n3 >= 1) ans = min(ans, dp(num-1, n1+2, n2, n3-1) + 1);
        return ans;
    }
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(a, -1, sizeof(a));
        int n1, n2, n3, n4;
        scanf("%d%d%d%d", &n1, &n2, &n3, &n4);
        printf("%d\n", dp(n1, n2, n3, n4));
    }
    return 0;
}