P1596 [USACO10OCT]湖计数Lake Counting
题目描述
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.
由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水('W') 或是旱地('.')。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。
输入输出格式
输入格式：
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.
第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是'W'或'.'，它们表示网格图中的一排。字符之间没有空格。
输出格式：
Line 1: The number of ponds in Farmer John's field.
一行：水坑的数量
输入输出样例
10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.
3
说明
OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left, and one along the right side.
#include#include #include #include #include #define N 110using namespace std;int n,m,ans,xx[8]={ 0,0,1,1,1,-1,-1,-1},yy[8]={ 1,-1,1,0,-1,1,0,-1};char ch[N][N];bool vis[N][N];int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1; ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();} return x*f;}void dfs(int x,int y){ if(x<0||y<0||x>n||y>m||vis[x][y]) return ; vis[x][y]=true; for(int i=0;i<8;i++) { if(ch[x+xx[i]][y+yy[i]]=='W') dfs(x+xx[i],y+yy[i]); }}int main(){ n=read(),m=read(); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>ch[i][j]; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(ch[i][j]=='W'&&!vis[i][j]) ans++,dfs(i,j); printf("%d",ans);}