广搜写法

两个队列分别表示两个坐标

#include <bits/stdc++.h>
using namespace std;
int n, m;
char g[115][115];
// 把与 (x,y) 连通的 'W' 都改为 '.'
int dx[8] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[8] = {1, -1, 0, 0, -1, 1, -1, 1};
queue<int> qx, qy;
void bfs(int x, int y)
{
    // 起点入队
    qx.push(x);
    qy.push(y);
    g[x][y] = '.';
    // 队列非空就继续搜
    while (!qx.empty())
    {
        // 取出队头
        int now_x = qx.front();
        int now_y = qy.front();
        qx.pop();
        qy.pop();
        // 考虑 now_x,now_y 能走到哪儿
        for (int i = 0; i < 8; i++)
        {
            // now_x,now_y 的第 i 个方向的点是 next_x,nextY
            int next_x = now_x + dx[i];
            int next_y = now_y + dy[i];
            // 能走到，且没走过
            if (1 <= next_x && next_x <= n &&
                1 <= next_y && next_y <= m &&
                g[next_x][next_y] == 'W')
            {
                qx.push(next_x);
                qy.push(next_y);
                g[next_x][next_y] = '.';
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> g[i][j];
    // 算水洼数量
    int ans = 0;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            if (g[i][j] == 'W')
            {
                ans++;
                bfs(i, j); // 把与 i,j 连通的 'W' 都改为 '.'
            }
        }
    cout << ans;
    return 0;
}

用 vis 标记有没有被统计过

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 110;
int n, m;
char a[MAXN + 5][MAXN + 5];
int dx[] = {-1, 1, 0, 0, -1, 1, -1, 1};
int dy[] = {0, 0, -1, 1, -1, -1, 1, 1};
// 每个位置有没有被统计过，false 表示没有被统计过
bool vis[MAXN + 5][MAXN + 5];
// 广搜的队列
queue<pair<int, int>> q;
// 把 (x,y) 相连的 W 全都标记为统计过了
void bfs(int x, int y)
{
    // 起点入队
    q.push(make_pair(x, y));
    vis[x][y] = true;
    // 只要还有东西要改，就改
    while (!q.empty())
    {
        // 取出队头
        pair<int, int> now = q.front();
        q.pop();
        int nowX = now.first;
        int nowY = now.second;
        // 检查当前位置八连通的所有位置
        for (int i = 0; i < 8; i++)
        {
            int nxtX = nowX + dx[i];
            int nxtY = nowY + dy[i];
            // 在范围内、是 W、没统计过，就继续搜
            if (1 <= nxtX && nxtX <= n &&
                1 <= nxtY && nxtY <= m &&
                a[nxtX][nxtY] == 'W' &&
                vis[nxtX][nxtY] == false)
            {
                q.push(make_pair(nxtX, nxtY));
                vis[nxtX][nxtY] = true;
            }
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];

    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (a[i][j] == 'W' && !vis[i][j])
            {
                ans++;
                // 把这个水洼全都标记为被算过了
                bfs(i, j);
            }
        }
    }
    cout << ans;
    return 0;
}

直接改成 .

#include <bits/stdc++.h>
using namespace std;
int n, m, ans;
char g[115][115];
//广搜
bool vis[115][115];
struct Point
{
    int x, y;
};
queue<Point> q;
int dx[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy[] = {1, -1, 1, -1, 0, 0, 1, -1};
void bfs(int sx, int sy)
{
    //初始化
    memset(vis, false, sizeof(vis));
    while (!q.empty())
        q.pop();
    //起点入队
    q.push((Point){sx, sy});
    vis[sx][sy] = true;
    while (!q.empty())
    {
        //取出队头
        Point now = q.front();
        q.pop();
        //枚举能到达的所有点
        for (int i = 0; i < 8; i++)
        {
            int nx = now.x + dx[i];
            int ny = now.y + dy[i];
            // (now.x, now.y) -> (nx, ny)
            if (1 <= nx && nx <= n &&
                1 <= ny && ny <= m &&
                !vis[nx][ny] &&
                g[nx][ny] == 'W')
            {
                q.push((Point){nx, ny});
                g[nx][ny] = '.';
                vis[nx][ny] = true;
            }
        }
    }
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> g[i][j];
    ans = 0;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (g[i][j] == 'W')
            {
                ans++;
                //把这个点对应的水洼抽干
                bfs(i, j);
            }
    cout << ans << endl;
    return 0;
}

深搜写法

#include<bits/stdc++.h>
using namespace std;
int n,m;
char g[115][115];
int dx[] = {-1,-1,-1,0,0,1,1,1};
int dy[] = {-1,0,1,-1,1,-1,0,1};
//把 (xy) 对应的联通块清空 
void dfs(int x,int y){
	g[x][y]='.';
	for(int i=0;i<8;i++){
		int nx=x+dx[i];
		int ny=y+dy[i];
		if(1<=nx&&nx<=n&&
		   1<=ny&&ny<=m&&
		   g[nx][ny]=='W'){
			dfs(nx,ny);   	
		}
	}	
}
int main()
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			cin>>g[i][j];
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(g[i][j]=='W'){
				ans++;
				dfs(i,j);
			}
	cout<<ans<<endl;
	return 0;
}