Hdu4800_josephina and RPG (two-dimensional state DP)

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The main puzzle: the status of DP is set to, using the role J to pass the maximum winning probability (Dp[i][j])

From the last start, Count:
DP[I][J] = rate[j][Ai[i]] * MAX (dp[i+1][j], dp[i+1][a[i]);

That is: Use the role J to go through the maximum winning rate of the I-pass = use role J to win the first I-off AI's winning rate * MAX (the probability of the next close with J clearance, the next level of the probability of the AI pass the switch)

Set the initial value, the calendar can be.

#include <sstream>#include <string>#include <iostream>#include <algorithm>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iterator>#include <vector>#include <map>#include <stack>#include <queue>#include <set>#include <list>#include <functional>#include <numeric>using namespace STD;#define MEM (x, y) memset (x,y,sizeof (×) )inline intLowbit (intx) {returnx& (-X);}typedef  Long Long intLL;Const intINF =0x3f3f3f3f;Const Long DoublePI =ACOs(0.0) *2.0;Const intN =10000+Ten;Const DoubleEPS =1e-6;Doubledp[n][ $],rate[ $][ $];intAi[n];intComb (intXinty);intMain () {intN,m;//ios::sync_with_stdio (FALSE);//close the synchronization stream     while(scanf("%d", &m) = =1)    {intR = Comb (m,3); for(inti =0; I < R; i++) for(intj =0; J < R; J + +)scanf("%LF", &rate[i][j]);scanf("%d", &n); for(inti =0; I < n; i++)scanf("%d", &ai[i]); MEM (DP,0.0); for(intj =0; J < R; J + +) dp[n-1][J] = rate[j][ai[n-1] ]; for(inti = n2; I >=0; i--) { for(intj =0; J < R; J + +) {Dp[i][j] = max (dp[i][j],rate[j][ai[i]] * MAX (dp[i+1][j],dp[i+1[Ai[i]])); }        }Doubleres =0.0; for(intj =0; J < R; J + +) res = max (res,dp[0][J]);printf("%.6lf\n", res); }return 0;}intComb (intXintY) {intu=1, v=1, Len;if(y==0|| X==y)return 1;if(y==1)returnXif(x < y)return 0;if(Y > x/2) y = len = x-y;ElseLen = y; while(len--)        {u*= x--;    V*= y--; }return(int) (u/v);}

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Hdu4800_josephina and RPG (two-dimensional state DP)