A - Lake Counting解题报告

A - Lake Counting

Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

这道题由于学长讲过原理，所以我就照着学长讲的做了。第一次做DFS，感觉还不错嘛。

期间由于宽高搞错互换，还有读数据的时候读入了回车导致数据不正确，这种问题还是蛮多的。

原理就是检索水洼，如果这个水洼没有被DFS过的话，水洼数加一，对它DFS，把周围的水洼都标记了，这样继续检索的时候就不会在计算入内了。

//Lake Counting

#include <stdio.h>

char Map[101][101];
int MapVisited[101][101] = {0};
int MapWidth,MapHeight;
int WaterNum;

void DFS(int x,int y);
char JudgeWater(int x,int y);

void main()
{
	int i,j;
	//Read Map Data
	scanf("%d %d",&MapHeight,&MapWidth);
	getchar();
	for (j = 1;j <= MapHeight;j ++)
	{
		for (i = 1;i <= MapWidth;i ++)
		{
			scanf("%c",&Map[i][j]);
		}
		getchar();
	}

	/*
	for (j = 1;j <= MapHeight;j ++)
	{
		for (i = 1;i <= MapWidth;i ++)
		{
			printf("%c",Map[i][j]);
		}
		putchar('\n');
	}
	*/

	WaterNum = 0;
	//DFS
	for (j = 1;j <= MapHeight;j ++)
	{
		for (i = 1;i <= MapWidth;i ++)
		{
			if (Map[i][j] == 'W' && MapVisited[i][j] == 0)//Not Searched Water
			{
				//printf("x %d y %d\n",i,j);
				WaterNum ++;
				DFS(i,j);
			}
		}
	}
	printf("%d\n",WaterNum);
}

void DFS(int x,int y)
{
	if (MapVisited[x][y] == 0)
	{
		MapVisited[x][y] = 1;
		//printf("Vistied x %d y %d\n",x,y);
		
		if (JudgeWater(x - 1,y - 1) == 'W')
		{
			DFS(x - 1,y - 1);
		}
		if (JudgeWater(x - 1,y) == 'W')
		{
			DFS(x - 1,y);
		}
		if (JudgeWater(x - 1,y + 1) == 'W')
		{
			DFS(x - 1,y + 1);
		}
		if (JudgeWater(x,y - 1) == 'W')
		{
			DFS(x,y - 1);
		}
		if (JudgeWater(x,y + 1) == 'W')
		{
			DFS(x,y + 1);
		}
		if (JudgeWater(x + 1,y - 1) == 'W')
		{
			DFS(x + 1,y - 1);
		}
		if (JudgeWater(x + 1,y) == 'W')
		{
			DFS(x + 1,y);
		}
		if (JudgeWater(x + 1,y + 1) == 'W')
		{
			DFS(x + 1,y + 1);
		}
	}
}

char JudgeWater(int x,int y)
{
	if (x >= 1 && x <= MapWidth && y >= 1 && y <= MapHeight)
	{
		return Map[x][y];
	}
	return '.';
}