HDU 4865 Peter's Hoby (probability DP)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4865

There are three kinds of weather and Four Leaf states. The two tables are given, respectively, the probability of leaf state in each weather and the probability of today's weather to tomorrow's weather. The status of the leaves in N days is given to output the most likely weather sequence.

Thought: WL [I] [J] indicates the probability that the weather is I and the leaves are J. ww [I] [J] indicates that the weather today is I. The probability that the weather tomorrow is J, st [I] indicates the probability that the weather on the first day is I.

For the leaf sequence {A1, a2 ...... an}, there is a weather sequence {B1, B2 ...... BN }, then the total probability G [N] = sT [B1] * WL [B1] [a1] * ww [B1] [B2] * WL [B2] [a2] *...... * ww [bn-1] [bn] * WL [bn] [an]. That is, log (G [N]) = Log (ST [B1]) + Log (WL [B1] [a1]) + Log (WW [B1] [B2]) + Log (WL [B2] [a2]) + ...... + Log (WW [bn-1] [bn]) + Log (WL [bn] [an]). The sequence corresponding to the maximum value of log (G [N]) is the weather sequence.

It is equivalent to giving a matrix of N * 3, representing n days. There are 3 kinds of weather every day. From the first line to the nth line, we can find a longest path and output path.

# Include <stdio. h> # include <iostream> # include <map> # include <set> # include <stack> # include <vector> # include <math. h> # include <string. h> # include <queue> # include <string> # include <stdlib. h >#include <algorithm> # define ll long # DEFINE _ ll _ int64 # define EPS 1e-12 # define PI ACOs (-1.0) using namespace STD; const int INF = 0x3f3f3f; double WL [3] [4] = {0.6, 0.2, 0.15, 0.05}, {0.25, 0.3, 0.2, 0.25}, {0.05, 0.10, 0. 35, 0.50 }}; double WW [3] [3] = {0.5, 0.375, 0.125}, {0.25, 0.125, 0.625}, {0.25, 0.375, 0.375 }}; double st [3] = {0.63, 0.17, 0.2}; int N; char s [10]; int A [55]; double f [55] [5]; double DP [55] [5]; int pre [55] [5]; stack <int> sta; int Init (char s []) {If (strcmp (S, "dry") = 0) return 0; If (strcmp (S, "dryish") = 0) return 1; if (strcmp (S, "damp") = 0) return 2; If (strcmp (S, "soggy") = 0) return 3 ;} void to () {for (INT I = 0; I <3; I ++) {For (Int J = 0; j <3; j ++) WW [I] [J] = Log (WW [I] [J]);} for (INT I = 0; I <3; I ++) ST [I] = Log (ST [I]);} int main () {(); int test; scanf ("% d", & Test); For (INT item = 1; item <= test; item ++) {scanf ("% d ", & N); memset (PRE,-1, sizeof (pre); memset (F, 0, sizeof (f); For (INT I = 1; I <= N; I ++) {scanf ("% s", S); A [I] = Init (s) ;}for (INT I = 1; I <= N; I ++) {// The leaf status on the current day is fixed, the current F [I] [J] indicates the probability that the weather on day I is J under fixed leaf conditions. // Therefore, F [I] [J] indicates each type of weather. Ratio of data to the total weather: Double T = 0; For (Int J = 0; j <3; j ++) {f [I] [J] = WL [J] [A [I]; t + = WL [J] [A [I];} for (Int J = 0; j <3; j ++) f [I] [J]/= T;} For (INT I = 1; I <= N; I ++) {for (Int J = 0; j <3; j ++) f [I] [J] = Log (F [I] [J]);} memset (DP,-INF, sizeof (DP); For (Int J = 0; j <3; j ++) DP [1] [J] = sT [J] + F [1] [J]; for (INT I = 2; I <= N; I ++) {for (Int J = 0; j <3; j ++) {for (int K = 0; k <3; k ++) {double T = DP [I-1] [k] + WW [k] [J] + F [I] [J]; If (DP [I] [J] <t) {DP [I] [J] = T; pre [I] [J] = K ;}}} int P = N; int C; double max =-INF; For (Int J = 0; j <3; j ++) {If (DP [N] [J]> MAX) {max = DP [N] [J]; C = J ;}} while (! Sta. Empty () Sta. Pop (); Sta. Push (c); While (pre [p] [c]! =-1) {Sta. push (pre [p] [c]); C = pre [p] [c]; p --;} printf ("case # % d: \ n", item ); while (! Sta. empty () {If (Sta. top () = 0) printf ("Sunny \ n"); If (Sta. top () = 1) printf ("Cloudy \ n"); If (Sta. top () = 2) printf ("Rainy \ n"); Sta. pop () ;}} return 0 ;}

HDU 4865 Peter's Hoby (probability DP)