广搜写法

#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;
}

#include<bits/stdc++.h>
using namespace std;
int n,m;
char g[115][115];
queue<int> qx,qy;
int dx[8] = {-1,-1,-1,0,0,1,1,1};
int dy[8] = {-1,0,1,-1,1,-1,0,1};
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++;
				//把所有与 (i,j) 相连的水全都去掉		
				//起点入队
				qx.push(i);
				qy.push(j);
				g[i][j]='.';
				//只要队列非空（还有需要处理的水）
				while(qx.size()>0){
					//取出队头 
					int nowX = qx.front();qx.pop();
					int nowY =qy.front();qy.pop(); 
					//枚举所有队头一步能走到的位置，如果没走过，就走一下
					for(int i=0;i<8;i++){
						int nxtX = nowX+dx[i];
						int nxtY = nowY+dy[i];
						if(1<=nxtX&&nxtX<=n&&1<=nxtY&&nxtY<=m&&g[nxtX][nxtY]=='W')
						{
							g[nxtX][nxtY]='.';
							qx.push(nxtX);
							qy.push(nxtY);
						}						
					} 
				}
				 
			}			
		} 
	cout<<ans<<"\n"; 
	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;
}