Time limit1000 ms
Memory limit65536 kB
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 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
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.
题意：W是水洼，连起来的W算同一个水洼（是九宫格内的连起来）
题解：dfs搜索，搜索不到了就继续，每一个dfs都可以搜到一个水坑，简而言之，总的dfs的次数就是水坑的个数（dfs重新调用的dfs不算）
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
using namespace std;
#define PI 3.14159265358979323846264338327950
int N,M;
const int MAX_N=103;
char field[MAX_N][MAX_N];
void dfs(int x,int y)
{
field[x][y]=‘.‘;
for(int dx=-1;dx<=1;dx++)
{
for(int dy=-1;dy<=1;dy++)
{
int nx=x+dx,ny=y+dy;
if(0<=nx && nx<N && 0<=ny && ny<M && field[nx][ny]==‘W‘)
dfs(nx,ny);
}
}
return ;
}
void solve()
{
int res =0;
for(int i=0;i<N;i++)
{
for(int j=0;j<M;j++)
{
if(field[i][j]==‘W‘)
{
dfs(i,j);
res++;
}
}
}
printf("%d\n",res);
}
int main()
{
cin>>N>>M;
for(int i=0;i<N;i++)
for(int j=0;j<M;j++)
cin>>field[i][j];
solve();
}
原文地址：https://www.cnblogs.com/smallhester/p/9499097.html