1359. Count All Valid Pickup and Delivery Options Hard
Given n orders, each order consist in pickup and delivery services.
Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i).
Since the answer may be too large, return it modulo 10^9 + 7.
Example 1:
Input: n = 1 Output: 1 Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.
Example 2:
Input: n = 2 Output: 6 Explanation: All possible orders: (P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1). This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.
Example 3:
Input: n = 3 Output: 90
Constraints:
1 <= n <= 500
class Solution {
public int countOrders(int n) {
long result = 0;
return (int)dfs(n, n, new long[n + 1][n + 1]);
}
private long dfs(int p, int d, long[][] mem) {
long mod = 1000000007;
//基本条件，当p和d位0的时候只有1中选择
if(p == 0 && d == 0) return 1;
//如果剩余pickup比delivery都多 不可能，负值也不可能
if(p > d || p < 0 || d < 0) return 0;
if(mem[p][d] > 0) return mem[p][d];
long count = 0;
//有p中方法取一个pickup，因此有： p * dfs(p - 1, d, mem)
count = (p * dfs(p - 1, d, mem)) % mod;
//如果pickup完成的多（剩余的少）， 有d-p个可以delivery 因此 d-p种， 因此有 (d - p) * dfs(p , d - 1) 种
if(p < d) count = (count + (d - p) * dfs(p , d - 1, mem)) % mod;
mem[p][d] = count;
return count;
}
}
标签：P2,P1,mem,memo,dfs,D2,D1 From： https://www.cnblogs.com/cynrjy/p/16935274.html