POJ

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 38395 Accepted: 19066

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

Sample Output

3

题意：计算出湖的数目；

思路：题目中指出是八个方向的搜索，用深搜枚举八个方向计算出湖的数目。

#include<stdio.h>char Map[110][110];//建图和标记两个功能int dir[8][2] = {0,1,1,1,1,0,1,-1,0,-1,-1,-1,-1,0,-1,1};//八个方向int r,c;void dfs(int x,int y){    int i;    for(i = 0; i < 8; i ++)//枚举八个方向    {        int tx = x + dir[i][0];        int ty = y + dir[i][1];        if(tx<r && tx>=0 && ty<c && ty>=0 && Map[tx][ty]=='W')        {            Map[tx][ty] = '.';//标记已搜过            dfs(tx,ty);        }    }}int main(){    int i,j,k;    while(~scanf("%d%d",&r,&c))    {        for(i = 0; i < r; i ++)            scanf("%s",Map[i]);        int sum = 0;        for(i = 0; i < r; i ++)        {            for(j = 0; j < c; j ++)            {                if(Map[i][j]=='W')//一趟搜索得到一个湖                {                    dfs(i,j);                    sum++;                }            }        }        printf("%d\n",sum);    }    return 0;}