Matrix Fetch Game NOIP 2007

2016-05-31 17:26:45

Title Link: NOIP 2007 matrix Fetch game (CODEVS)

Main topic:

Given a matrix, each time at the beginning of each line or at the end of the line takes a number multiplied by the number of 2^, to get the most points obtained

Solution:

Dynamic planning

DP[I][J] Indicates the maximum score obtained from the current line I position to the J position

Transfer equation:

Dp[i][j]=max (DP[I+1][J]+A[I]*2^X,DP[I][J-1]+A[I]*2^X);

Places to be aware of:

1.M and N give the range of a little pit ah, 2^80 only quietly do not speak, so directly on the 100000 of the pressure, feel more than better than fine

2.2 of the time to use, so pre-processing out to use

1 //Matrix Fetch Game (NOIP2007)2 //dynamic programming with high accuracy3#include <stdio.h>4#include <algorithm>5#include <string.h>6 using namespacestd;7 Const intmaxn= -;8 Const intmaxm= -;9 Const intMaxl= -;Ten intn,m; One intA[MAXN]; A intDP[MAXN][MAXM][MAXL]; - intYS[MAXN][MAXL]; - intANS[MAXL]; the intTMP[MAXL]; - voidMulti (int*x,int*y,intz) - { -memset (TMP,0,sizeof(TMP)); +tmp[0]=y[0]; -      for(intI=1; i<=tmp[0];i++) +     { Atmp[i]=y[i]*Z; at     } -      for(intI=1; i<=tmp[0];i++) -     { -tmp[i+1]+=tmp[i]/100000; -tmp[i]%=100000; -     } in      while(tmp[tmp[0]+1]) -     { totmp[0]++; +tmp[tmp[0]+1]+=tmp[tmp[0]]/100000; -tmp[tmp[0]]%=100000; the     } *      for(intI=0;i< -; i++) x[i]=Tmp[i]; $     return ;Panax Notoginseng } - voidADD (int*x,int*y,int*z) the { +memset (TMP,0,sizeof(TMP)); A     if(y[0]>z[0]) tmp[0]=y[0]; the     Elsetmp[0]=z[0]; +      for(intI=1; i<=tmp[0];i++) -     { $tmp[i]=y[i]+Z[i]; $     } -      for(intI=1; i<=tmp[0];i++) -     { thetmp[i+1]+=tmp[i]/100000; -tmp[i]%=100000;Wuyi     } the     if(tmp[tmp[0]+1]) tmp[0]++; -      for(intI=0;i< -; i++) x[i]=Tmp[i]; Wu     return ; - } About BOOLMax (int*x,int*y) $ { -     if(x[0]>y[0])return 1; -     Else if(y[0]>x[0])return 0; -     Else  A     { +          for(inti=x[0];i>=1; i--) the         { -             if(X[i]>y[i])return 1; $             Else if(X[i]<y[i])return 0; the         } the     } the     return 1; the } - intMain () in { thescanf"%d%d",&n,&M); theys[0][0]=1; Aboutys[0][1]=1; the      for(intI=1; i<=m;i++) the     { theMulti (ys[i],ys[i-1],2); +     } -ans[0]=1; ans[1]=0; the      for(intI=1; i<=n;i++)Bayi     { the          for(intj=1; j<=m;j++) the         { -scanf"%d",&a[j]); - Multi (Dp[j][j],ys[m],a[j]); the         } the          for(intk=2; k<=m;k++) the         { the              for(intx=1; x<=m-k+1; x + +) -             { the                 inty=x+k-1; the                 intxm[ to],ym[ to],cm[ to]; theMulti (cm,ys[m-k+1],a[x]);94ADD (xm,cm,dp[x+1][y]); theMulti (cm,ys[m-k+1],a[y]); theADD (ym,cm,dp[x][y-1]); the                 if(Max (xm,ym))98                 { Aboutmemcpy (DP[X][Y],XM,sizeof(int)* -); -                 }101                 Else 102                 {103memcpy (Dp[x][y],ym,sizeof(int)* -);104                 } the             }106         }107ADD (ans,ans,dp[1][m]);108     }109printf"%d", ans[ans[0]]); the      for(inti=ans[0]-1; i>=1; i--)111     { theprintf"%05d", Ans[i]);113     } theprintf"\ n"); the     return 0; the}

Matrix Fetch Game NOIP 2007