Codeforces431c-K-tree DP

Q: I will give you a K-Cross number. Each node must have k subnodes, and the cost for these K subnodes is 1-K, ask you the probability that the total cost is N and at least one D-K edge needs to be crossed

Solution: divide the problem into two-dimensional DP, one indicating that the process has not passed, and the other indicates that the process has passed, and the DP can be used to find the solution.

Solution code:

1 /************************************** * ********************** 2 * Author: darkdream 3 * Email: [email protected] 4 * Last modified: 5 * filename: 431c. CPP 6 * description: 7 *************************************** * ******************/8 // File Name: 431c. CPP 9 // Author: darkdream10 // created time: friday August 01, 2014 11 12 # include <vector> 13 # include <list> 14 # include <map> 15 # include <set> 16 # include <deque> 17 # include <stack> 18 # include <bitset> 19 # include <algorithm> 20 # include <functional> 21 # include <numeric> 22 # include <utility> 23 # include <sstream> 24 # include <iostream> 25 # include <iomanip> 26 # include <cstdio> 27 # include <cmath> 28 # include <cstdlib> 29 # include <cstring> 30 # include <ctime> 31 32 # define ll long long33 using namespace STD; 34 ll DP [104] [104]; 35 ll dpa [104] [104]; 36 # define MD 100000000737 int main () {38 int N, K, D; 39 scanf ("% d", & N, & K, & D); 40 memset (DP, 0, sizeof (DP); 41 memset (DPA, 0, sizeof (DPA); 42 DP [0] [0] = 1; 43 ll ans = 0; 44 for (INT I = 1; I <= N; I ++) 45 {46 for (Int J = I-1; j <= N; j ++) 47 {48 if (DP [I-1] [J]) 49 {50 for (int s = 1; S <= K; s ++) 51 {52 If (J + S> N) 53 break; 54 if (S <D) 55 DP [I] [J + S] = (DP [I] [J + S] + dp [I-1] [J]) % md; 56 else 57 DPA [I] [J + S] = (DPA [I] [J + S] + dp [I-1] [J]) % md; 58} 59} 60 if (DPA [I-1] [J]) 61 {62 for (int s = 1; S <= K; s ++) 63 {64 if (J + S> N) 65 break; 66 DPA [I] [J + S] = (DPA [I-1] [J] + DPA [I] [J + S]) % md; 67} 68} 69} 70} 71 for (INT I = 1; I <= N; I ++) 72 ans = (DPA [I] [N] + ans) % md; 73 printf ("% i64d \ n", ANS % MD); 74 return 0; 75}

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