DP in the HDU 5001 probability DP Diagram

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 5001

At that time, when I saw the figure, I fell down and did not dare to write it, nor did I return the DP equation.

I feel that the question of the semi-finals is a bit difficult to think of. The second is that coding is a bit difficult.

This question:

I keep thinking that 1-the probability of reaching I is the probability of not reaching I, and then the three-dimensional equation barabara, confused myself.

Correct practice:
DP [k] [J] After J steps reach K points and the probability of not passing through I points is set so that the probability of not reaching I points can be obtained. Segma (DP [v] [J-1]/g [v]. size () V is the adjacent node with K, if the J-1 step reaches step v j there is 1/g [v]. the probability of size () reaches K.

// # Pragma comment (linker, "/Stack: 102400000,102400000 ") # include <cstdio> # include <cstring> # include <algorithm> # include <string> # include <iostream> # include <iomanip> # include <cmath> # include <map> # include <set> # include <queue> using namespace STD; # define ls (RT) rt * 2 # define RS (RT) rt * 2 + 1 # define ll long # define ull unsigned long # define rep (I, S, e) for (INT I = s; I <E; I ++) # define repe (I, S, e) for (int I = s; I <= E; I ++) # define Cl (a, B) memset (a, B, sizeof (A) # define in (s) freopen (S, "r", stdin) # define out (s) freopen (S, "W", stdout) const ll ll_inf = (ull) (-1)> 1; const double EPS = 1e-8; const double Pi = ACOs (-1.0); const int INF = 100000000; const int maxn = 60; const int max = 10000 + 100; double ans [maxn]; double DP [maxn] [Max]; vector <int> G [Max]; int n, m, D; void solve () {for (INT I = 1; I <= N; I ++) {Cl (DP, 0); For (int J = 1; j <= D; j ++) {If (j = 1) {for (int K = 1; k <= N; k ++) if (K! = I) DP [k] [J] = 1.0/N; // select a random vertex in the first step to reach any vertex, so the probability is 1.0/n} else {for (int K = 1; k <= N; k ++) if (K! = I) {for (INT T = 0; t <G [K]. size (); t ++) {int v = G [k] [T]; If (V! = I) DP [k] [J] + = DP [v] [J-1]/g [v]. size () ;}}} double out = 0.0; For (Int J = 1; j <= N; j ++) if (J! = I) out + = DP [J] [d]; printf ("%. 10lf \ n ", out) ;}} int main () {// In (" hdu5001.txt "); int ncase; scanf (" % d ", & ncase ); while (ncase --) {for (INT I = 0; I <= N; I ++) g [I]. clear (); // For (INT I = 0; I <= N; I ++) ans [I] = 0.0; scanf ("% d ", & N, & M, & D); D ++; // int U, V; For (INT I = 0; I <m; I ++) {scanf ("% d", & U, & V); G [u]. push_back (V); G [v]. push_back (U);} solve ();} return 0 ;}

DP in the HDU 5001 probability DP Diagram