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
输入 #1
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
输出 #1
3
说明/提示
OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left, and one along the right side.
思路：BFS即可
代码如下
#include<bits/stdc++.h>
using namespace std;
int u[8]={1,1,0,-1,-1,-1,0,1},v[8]={0,1,1,1,0,-1,-1,-1};
char a[101][101];
int flag[101][101];
int ans,m,n,cnt=1;
//这题的意思是只要有w就能成一个水坑 只要周围也有w 也把它也算到一个水坑里面
struct A{int x,y;
}s;
void bfs(int xx,int yy){queue<A> q;A o;o.x=xx;o.y=yy;q.push(o);//借助结构体存储一个点的坐标 flag[xx][yy]=cnt;//每个水坑cnt的数值不一样，因此能使水坑有所区分 while(!q.empty()){o=q.front();//队头 q.pop();//出队 for(int i=0;i<8;i++){int nx=o.x+u[i];int ny=o.y+v[i];if(nx<1||ny<1||nx>n||ny>m||flag[nx][ny]!=0||a[nx][ny]!='W'){continue;}flag[nx][ny]=cnt;A next;next.x=nx;next.y=ny;q.push(next);}}
}
int main(){cin>>n>>m;for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){cin>>a[i][j];}for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){if(a[i][j]=='W'&&flag[i][j]==0){bfs(i,j);ans++;cnt++;//这里的++可以使不同水坑有不同的标记}}cout<<ans;return 0;
}记录一下 回看