DFS模板题---Lake Counting

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

Status

Practice

POJ 2386
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
  
  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.

标准的DFS模板题目，代码如下：

#include <stdio.h>
#include <string.h>
#define lim 500
void dfs(int x, int y, int map[][lim], int vis[][lim]);
int main ()
{
    int row, col;
    while(scanf("%d%d", &row, &col)!=EOF)
    {
        if(row==0)
            return 0;
        int i, j, count=0;
        char str[800];
        int  map[500][500], vis[500][500];//定义两个数组
        memset(str, 0, sizeof(str));
        memset(map, 0, sizeof(map));
        memset(vis, 0, sizeof(vis));
        for(i=0;i<row;i++)//进行题目中的输入
        {
           scanf("%s", str);//输入方法
           for(j=0;j<col;j++)
                map[i+1][j+1]=str[j]-'.';//相当于外边加一层 * ，防止搜索时数组越界！！！
        }
        for(i=1;i<=row;i++)
            for(j=1;j<=col;j++)
            if(map[i][j]==41 && vis[i][j]==0)//搜索时遇到还未被访问的@则进行搜索
        {
            dfs(i, j, map, vis);
            count++;
        }
        printf("%d\n", count);
    }
    return 0;
}

void dfs(int x, int y, int map[][lim], int vis[][lim])
{
    if(map[x][y]==0||vis[x][y]==1)
        return;
     vis[x][y]=1;
    dfs(x, y-1, map, vis );//上
    dfs(x, y+1, map, vis );//下
    dfs(x-1, y , map, vis);//左
    dfs(x+1, y, map, vis);//右
    dfs(x-1, y-1, map, vis);//左上
    dfs(x-1, y+1, map, vis );//左下
    dfs(x+1, y-1, map, vis );//右上
    dfs(x+1, y+1, map, vis );//右下
}

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