#include<bits/stdc++.h>
using namespace std;
int n, m, ans;
int dx[] = {1, -1, 0, 0};//方向数组 
int dy[] = {0, 0, 1, -1}; 
char a[101][101];
void dfs(int x, int y)
{
	a[x][y] = '.';//已经走过
	int nx, ny;
	for(int i = 0; i < 4; i ++)
		for(int j = 0; j < 4; j ++)
		{
			nx = x + dx[i];//四处查找 
			ny = y + dy[i];
			if(nx >= 0 && x <= n && ny >= 0 && y <= m && a[nx][ny] == 'W')//判断边界 
				dfs(nx, ny);//回溯搜索 
		}
}
int main()
{
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	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')//查找到就搜 
			{
				dfs(i, j);
				ans ++;//答案加一 
			}
	cout << ans << "\n";
	return 0;
}

很好奇下面根据第一篇题解写的代码为啥答案就对了：

#include<bits/stdc++.h>
using namespace std;
int n, m, ans;
int dx[] = {1, -1, 0, 0};//方向数组 
int dy[] = {0, 0, 1, -1}; 
char a[101][101];
void dfs(int x, int y)
{
	a[x][y] = '.';//已经走过
	int nx, ny;
	for(int i = -1; i <= 1; i ++)//就从这里开始是看着题解写的
		for(int j = -1; j <= 1; j ++)
		{
			nx = x + i;//四处查找（这里也是）
			ny = y + j;
			if(nx >= 0 && x <= n && ny >= 0 && y <= m && a[nx][ny] == 'W')//判断边界（后面的和原来一样） 
				dfs(nx, ny);//回溯搜索 
		}
}
int main()
{
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	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')//查找到就搜 
			{
				dfs(i, j);
				ans ++;//答案加一 
			}
	cout << ans << "\n";
	return 0;
}