* 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.
* Line 1: The number of ponds in Farmer John‘s field.
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
3
数出农田里面水洼，w为水洼，.为陆地，w周围8格内若有水洼，这这些水洼是连通的，只算做一个。
分析：
白书模板题。用dfs来查找8格内的水洼，找到则再dfs查找该水洼周围8格是否还有连通。每连通在一起的大水洼记为1个。
代码：
#include<stdio.h> char a[103][103]; int N,M; int dfs(int y,int x) { a[y][x]=‘B‘; if(y<0||y>=N||x<0||x>=M) return 0; for(int dy=-1;dy<=1;dy++) { for(int dx=-1;dx<=1;dx++) { if(a[y+dy][x+dx]==‘W‘) { dfs(y+dy,x+dx); } } } return 0; } ? int main() { int res=0; scanf("%d %d",&N,&M); getchar(); for(int i=0;i<N;i++) { for(int j=0;j<M;j++) { a[i][j]=getchar(); } getchar(); } for(int i=0;i<N;i++) { for(int j=0;j<M;j++) if(a[i][j]==‘W‘) { dfs(i,j); res++; } } printf("%d\n",res); return 0; }