Sicily 1137 Riverbed (two-part rule)

"Computer algorithm design and analysis" gnawing book ...

A little to see the book, just to brush a water problem, just began to read the wrong question also.

　　　Note : The difference between all measurement points is not greater than Di and not adjacent two points ...

1 //1137. Riverbed2#include <iostream>3 using namespacestd;4 5 intMax_length (intD[],intLintRintm) {6     //calculates the maximum length and returns7     //D is an array that holds the height8     //L is subscript left boundary9     //R is subscript right boundaryTen     //m for maximum height difference One     if(L==R)return 1; A     if(l==r-1) { -         if((d[l]-d[r]<=m) && (d[l]-d[r]>=-m))return 2; -         Else return 1; the     } -     //two points -     intMid = (l+r)/2; -     intLM =max_length (d,l,mid,m); +     intRM = Max_length (d,mid+1, r,m); -     intMM =1; +     intmin, Max; AMin = max =D[mid]; at      for(inti = mid-1; I >= l; i--) { -         if((d[i]>=max-m) && (d[i]<=min+m)) { -++mm; -             if(d[i]<min) min =D[i]; -             Else if(d[i]>max) max =D[i]; -         } in         Else  Break; -     } to      for(inti = mid+1; I <= R; i++) { +         if((d[i]>=max-m) && (d[i]<=min+m)) { -++mm; the             if(d[i]<min) min =D[i]; *             Else if(d[i]>max) max =D[i]; $         }Panax Notoginseng         Else  Break; -     } the     //returns the maximum value +     return(LM&GT;RM?LM:RM) >mm? (lm>rm?)lm:rm): mm; A } the  + intMain () { -     intN, M; $Cin>>n>>m; $     intD[n]; -      for(inti =0; I < n; i++) cin>>D[i]; -Cout<<max_length (D,0, N-1, m) <<Endl; the     return 0; -}

I do not know whether I write a little trouble ...

Sicily 1137 Riverbed (two-part rule)