Number of puddles algorithm code (C)

Title: There is a garden of n*m size, the rain has accumulated water. The eight-connected stagnant water is considered to be connected together. Ask how many puddles there are in the garden altogether.

Use depth First search (DFS), and in a puddle, look in 8 directions until all the connected water is found. Specify the next puddle again until there is no puddle.

Then all the depth-first searches are the number of puddles. Time complexity O (8*m*n) =o (m*n).

Code:

* * * * * main.cpp * * * Created on:2014.7.12 * This column more wonderful content: HTTP://WWW.BIANCENG.CNHTTP://WWW.BIANCENG.CN/PROGRAMMING/SJJG /* author:spike * * #include <stdio.h> #include <stdlib.h> #include <string.h> #in  
    Clude <math.h> class Program {static const int max_n=20, MAX_M=20;  
    int N = ten, M = 12; Char field[max_n][max_m+1] = {"W ..... WW. ",". Www..... WWW "," .... Ww... WW. "," ..... WW. "," ..... W.. ",". W...... W.. ",". W.w ..... WW. "," W.W.W ... W. ",". W.W ... W. ",". W.......  
    W. "};  
        void Dfs (int x, int y) {field[x][y] = '. ';  for (int dx =-1; DX <= 1; dx++) {for (int dy =-1; dy <= 1; dy++) {int NX = X+DX,  
                NY = Y+dy; if (0<=dx&&nx<n&&0<=ny&&ny<=m&&field[nx][ny]== ' W ') DFS (NX, NY);  
    } return;  
        } public:void Solve () {int res=0;  
                    for (int i=0; i<n; i++) {for (int j=0; j<m; J + +) {if (field[i][j] = = ' W ') {  
                    DFS (I,J);  
                res++;  
    } printf ("result =%d\n", res);  
      
      
}  
};  
    int main (void) {program P;  
    P.solve ();  
return 0; }

Output:

result = 3

Author: csdn Blog Mystra