編程演算法 - 水窪的數量 代碼(C)

標籤：mystra   編程演算法   水窪的數量   深度優先搜尋   c   

水窪的數量 代碼(C)

本文地址: http://blog.csdn.net/caroline_wendy

題目: 有一個大小為N*M的園子, 雨後起了積水. 八連通的積水被認為是串連在一起的. 請求出園子裡總共有多少水窪.

使用深度優先搜尋(DFS), 在某一處水窪, 從8個方向尋找, 直到找到所有連通的積水. 再次指定下一個水窪, 直到沒有水窪為止.

則所有的深度優先搜尋的次數, 就是水窪數. 時間複雜度O(8*M*N)=O(M*N).

代碼:

 

/* * main.cpp * *  Created on: 2014.7.12 *      Author: spike */#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>class Program {static const int MAX_N=20, MAX_M=20;int N = 10, 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;}

輸出:

result = 3