Challenge Program 2.1.4 Exhaustion Search >> Depth First search

{' W ', '. ', '. ', ' w ', '. '},{' W ', '. ', '. ', '. ', '. ', '. ', '. ', '. ', '},{', '. ', '},{', '. ', '. ', '. ', '. ', '. ', '. ', '. ', '. ', '. ', ' W ' example, where there are 5 of stagnant water

title: garden with N*m lattice, w indicates stagnant water, '. ' Said dry ground, the water near the 8 areas as long as there is water, then these two connections, and counted as the same water, asked a total number of water?

idea: exhaustion Search, O (n*m), search for stagnant water becomes '. ', and search for connected water to turn it into '. ' Until this water search is over, and a+=1, where the water is all changed to '. '

1#include <stdio.h>2 voidDfsintXinty);3 voidsolve ();4     intn=5, m=5;5         Charch[5][5]={6{'W','.','.','W','.'},7{'W','.','.','.','.'},8{'.','.','.','W','.'},9{'W','.','.','.','.'},Ten{'W','.','W','W','W'} One                       }; A intMain () - { -  the solve (); -     return 0; - } -  + voidDfsintXinty) - { +ch[x][y]='.'; A      for(intx1=-1; x1<=1; x1++) at          for(inty1=-1; y1<=1; y1++) -     { -         intnx=x+x1,ny=y+Y1; -         if(nx>=0&&nx<n&&ny>=0&&ny<m&&ch[nx][ny]=='W')//nx,ny cannot go beyond the garden - DFS (nx,ny); -     } in     return ; - } to  + voidSolve () - { the     intA=0; *      for(intI=0; i<n;i++) $          for(intk=0; k<m;k++)Panax Notoginseng         if(ch[i][k]=='W') -         { the DFS (i,k); +a++; A         } theprintf"%d\n", a); +}

The code for 5*5,5 blocks of stagnant water is shown here.

Tips: Depth-First search for DFS, starting at the very beginning, traversing all possibilities, manipulating or enumerating them.

Challenge Program 2.1.4 Exhaustion Search >> Depth First search