879. Profitable Schemes

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 1:

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.

Example 2:

Input: G = 10, P = 5, group = [2,3,5], profit = [6,7,8]
Output: 7
Explanation: 
To make a profit of at least 5, the gang could commit any crimes, as long as they commit one.
There are 7 possible schemes: (0), (1), (2), (0,1), (0,2), (1,2), and (0,1,2).

Note:

1. 1 <= G <= 100
2. 0 <= P <= 100
3. 1 <= group[i] <= 100
4. 0 <= profit[i] <= 100
5. 1 <= group.length = profit.length <= 100

DP，dp[i][j][k]表示到第i个group为止（这样定义可以吧后向遍历转化成前向遍历），有j个人可用，还需要至少k利润的选择总数

Python容易TLE

class Solution {
    public int profitableSchemes(int G, int P, int[] group, int[] profit) {
    	int n = group.length;
        int[][][]dp=new int[n][G+1][P+1];
        for(int j=group[0]; j<=G; j++)
        	for(int k=0; k<=Math.min(profit[0], P); k++)
        		dp[0][j][k] += 1;
        for(int j=0; j<=G; j++)
        	dp[0][j][0] += 1;
        
        for(int i=1; i<group.length; i++)
        	for(int j=0;j <=G; j++)
        		for(int k=0;k<=P;k++) {
        			dp[i][j][k] = dp[i-1][j][k];
        			if(j>=group[i]) dp[i][j][k]+=dp[i-1][j-group[i]][Math.max(k-profit[i], 0)];
        			dp[i][j][k] %= 1000000007;
        		}
        return dp[n-1][G][P]%1000000007;
    }
}