Hdu--4800--dp

There's a total of 2 different states of transition.

When we move the state from X to Y, we can choose to change players or not to change players, but one thing to be aware of is that if you want to change players, it's just the latest game.

And there's always an increase in the number of matches. We always from (x-1)---->x It is also logical that we must first play in the X-1 field in order to play in the X game.

Dp[x,y] indicates that at the time of the current X game, I was numbering in Y for < attention because this is where the game must be done in order, so that means that the first n field is all the games are finished >

Dp[x][y] = max (Dp[x][y], dp[x-1][y]*p[y][team[x]);//No replacement

Dp[x][y] = max (dp[x][team[x], dp[x-1][y]*p[y][team[x]]);//replacement

1#include <iostream>2#include <algorithm>3 using namespacestd;4 5 intN, M;6 Const intSize = -;7 Doubledp[10010][size];8 DoubleP[size][size];9 intteam[10010];Ten  One voidSolve () A { -      for(inti =0; I<=m; i++ ) -dp[0][i] =1.0; the      for(inti =1; I<=n; i++ ) -     { -          for(intj =0; J<m; J + + ) -         { +DP[I][J] = max (Dp[i][j], dp[i-1][j]*p[j][Team[i]]); -dp[i][Team[i]] = max (dp[i][team[i]), dp[i-1][j]*p[j][Team[i]]); +         } A     } at } -  - intMain () - { -     Doubleans; -      while(~SCANF ("%d",&m)) in     { -m = m * (M-1) * (M-2) /6; to          for(inti =0; I<m; i++ ) +         { -              for(intj =0; J<m; J + + ) the             { *scanf"%LF",&p[i][j]); $             }Panax Notoginseng         } -scanf"%d",&n); the          for(inti =1; I<=n; i++ ) +         { Ascanf"%d",&team[i]); the         } +Memset (DP,0,sizeof(DP)); - solve (); $Ans =0; $          for(inti =0; I<m; i++ ) -         { -Ans =Max (ans, dp[n][i]); the         } -printf"%.6lf\n", ans);Wuyi     } the     return 0; -}

View Code

Today

It sucks.

Ffffffffk

　　

Hdu--4800--dp