编程算法 - 水洼的数量 代码(C)

Posted 2020-09-25 liguangsunls

水洼的数量 代码(C)

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

题目: 有一个大小为N*M的园子, 雨后起了积水. 八连通的积水被觉得是连接在一起的. 请求出园子里总共同拥有多少水洼.

使用深度优先搜索(DFS), 在某一处水洼, 从8个方向查找, 直到找到全部连通的积水. 再次指定下一个水洼, 直到没有水洼为止.

则全部的深度优先搜索的次数, 就是水洼数. 时间复杂度O(8*M*N)=O(M*N).

代码:

 

/*
 * 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;
}

输出:

result = 3