hackerrank# Hexagonal Grid

Original title Address

Tiling variants, the practice is similar

Assuming the floor is long below this, gray is unable to fill the void, initially can be n blocks outside the floor filled with gray, easy to handle the boundary

Suppose now that a part has been processed from the back forward, the green brick represents already traversed, and the blue dashed box represents the sub-problem that has been calculated

Now you want to traverse the red border tiles

There are only two possible types of paving methods:

If you can shop down, it's easy to push to another sub-problem

If it's right, it's going to be a bit of a hassle, two possibilities.

The first possibility, there is a block of gray bricks, as shown, then the statute becomes a sub-problem

The second possibility, no gray bricks, below can also be placed sideways, then the statute into another sub-question

After finishing this step, you can continue to work on other bricks, as well as similar steps

Code:

1#include <cmath>2#include <cstdio>3#include <vector>4#include <iostream>5#include <algorithm>6#include <cstring>7 using namespacestd;8 9 #defineMax_n 128Ten  One intN, T; A BOOLCan[max_n][max_n]; - BOOLU[max_n]; - BOOLD[max_n]; the  - intMain () { -     /*Enter your code here. Read input from STDIN. Print output to STDOUT*/ -CIN >>T; +      while(t--) { -CIN >>N; +  Amemset (U,0,sizeof(U)); atmemset (D,0,sizeof(d)); -          for(inti =0; i < N; i++) { -             CharC; -CIN >>C; -U[i] = c = ='0'?true:false; -         } in          for(inti =0; i < N; i++) { -             CharC; toCIN >>C; +D[i] = c = ='0'?true:false; -         } the  *Memset (CAN,0,sizeof(Can)); $          for(inti =0; I <2; i++)Panax Notoginseng              for(intj =0; J <2; J + +) -Can[n + i][n + j] =true; the  +         inti =N; A         intj =N; the          while(I >=0|| J >=0) { +             if(--j >=0) { -                 if(!D[j]) { $CAN[I][J] = can[i][j +1]; $}Else { -                     if(U[i]) -CAN[I][J] |= Can[i +1][j +1]; the                     if(D[j +1]) { -                         if(!U[i])WuyiCAN[I][J] |= Can[i +1][j +2]; the                         if(U[i] && u[i +1]) -CAN[I][J] |= Can[i +2][j +2]; Wu                     } -                 } About             } $             if(-I. >=0) { -                 if(!U[i]) { -CAN[I][J] = can[i +1][j]; -}Else { A                     if(D[j]) +CAN[I][J] |= Can[i +1][j +1]; the                     if(U[i +1]) { -                         if(!D[j]) $CAN[I][J] |= Can[i +2][j +1]; the                         if(D[j] && d[j +1]) theCAN[I][J] |= Can[i +2][j +2]; the                     } the                 } -             } in         } the  thecout << (can[0][0] ?"YES":"NO") <<Endl; About     } the     return 0; the}

hackerrank# Hexagonal Grid