Leetcode: 879. Profitable Schemes

Posted 2020-12-14 tmortred

Description

There are G people in a gang, and a list of various crimes they could commit.

The i-th crime generates a profit[i] and requires group[i] gang members to participate.

If a gang member participates in one crime, that member can‘t participate in another crime.

Let‘s call a profitable scheme any subset of these crimes that generates at least P profit, and the total number of gang members participating in that subset of crimes is at most G.

How many schemes can be chosen?  Since the answer may be very large, return it modulo 10^9 + 7.

Example

Input: G = 5, P = 3, group = [2,2], profit = [2,3]
Output: 2
Explanation: 
To make a profit of at least 3, the gang could either commit crimes 0 and 1, or just crime 1.
In total, there are 2 schemes.

Note

1 <= G <= 100
0 <= P <= 100
1 <= group[i] <= 100
0 <= profit[i] <= 100
1 <= group.length = profit.length <= 100

分析

    这道题目虽然是 hard 级别，但是比一般 medium 级别题目要简单。就是一般的 二维 dp，即 dp[g][p]. 关键的优化点是  在 p 大于等于 P 时，不在细分 p 

code

    def profitableSchemes(self, G, P, g, p):
        """
        :type G: int
        :type P: int
        :type group: List[int]
        :type profit: List[int]
        :rtype: int
        """
        dp = [[0 for _ in range(P+1)] for _ in range(G+1)]

        for i, _ in enumerate(p):
            if g[i] > G:
                continue
            for j in range(G, g[i]-1, -1):
                if p[i] >= P:
                    dp[j][P] += sum(dp[j-g[i]])
                else:
                    dp[j][P] += sum(dp[j-g[i]][P-p[i]:])
                for k in range(p[i], P):
                    dp[j][k] += dp[j-g[i]][k-p[i]]

            if p[i] >= P:
                dp[g[i]][P] += 1
            else:
                dp[g[i]][p[i]] += 1

        return sum([d[P] for d in dp])%1000000007
    

总结

Runtime: 668 ms, faster than 100.00% of Python online submissions for Profitable Schemes.
Memory Usage: 13.1 MB, less than 75.00% of Python online submissions for Profitable Schemes.
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