Lake Counting--poj2386

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23950		Accepted: 12099

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



这个题可以用深搜也可以用广搜，我就是从这个题，明白了两种搜索方式的不同

大家来体会一下吧！





DFS版：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

char map[101][101];
int mov[8][2]={0,1,0,-1,-1,0,1,0,1,1,-1,-1,1,-1,-1,1};
int m,n;
bool can(int x ,int y)//判断这个点能不能走
{
    if(x<0||x>m-1||y<0||y>n-1||map[x][y]=='.')
    return false;
    return true;
}

void dfs(int x,int y)
{
    int i,xx,yy;
    for(i=0;i<8;i++)
    {
        xx=x+mov[i][0];
        yy=y+mov[i][1];
        if(can(xx,yy))
        {
            map[xx][yy]='.';//走一个点就标记一下
            dfs(xx,yy);
        }
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        int sum=0;
        for(i=0;i<m;i++)
        scanf("%s",map[i]);
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(map[i][j]=='W')
                {
                    map[i][j]='.';
                    dfs(i,j);//每次进入搜索函数就把这个点周围能走的点走完
                    sum++;
                }
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

 

BFS版：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
char map[101][101];
int m,n;
int mov[8][2]={0,1,0,-1,1,0,-1,0,1,1,-1,-1,1,-1,-1,1};
struct node
{
    int a,b;
}ta,tb;//定义一个结构体用来存坐标
bool can(node x)
{
    if(x.a<0||x.a>m-1||x.b<0||x.b>n-1||map[x.a][x.b]=='.')
    return false;
    return true;
}


void bfs(int x,int y)
{
    queue<node> q;
    ta.a=x;
    ta.b=y;
    q.push(ta);//入队
    while(!q.empty())//直到把队列能访问的都访问过
    {
        int i;
        ta=q.front();
        q.pop();
        for(i=0;i<8;i++)
        {
            tb.a=ta.a+mov[i][0];
            tb.b=ta.b+mov[i][1];
            if(can(tb))
            {
                map[tb.a][tb.b]='.';
                q.push(tb);//如果可以访问就入队
            }
        }
    }
}


int main()
{
    int i,j;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        int sum=0;
        for(i=0;i<m;i++)
        scanf("%s",map[i]);
        for(i=0;i<m;i++)
        for(j=0;j<n;j++)
        {
            if(map[i][j]=='W')
            {
                map[i][j]='.';
                bfs(i,j);
                sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}