set，vector都可以 暂时存走了哪些点都可以都可 然后把走过的点赋值上ans if(jz[x][y]==0)判断走没走过优化避免重复计算值 其他的就是vector用法

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
ll n,m,ans;
char a[1005][1005];
bool pd[1005][1005]; 
ll jz[1005][1005];
vector<pair<ll,ll>>b;
void dfs(int x,int y,char syg)
{
	if(jz[x][y]!=0||pd[x][y]==true|| x<1||x>n|| y<1||y>n|| syg==a[x][y])
		return;
	pd[x][y]=true;
	ans++;
  b.push_back(make_pair(x,y));
	dfs(x+1,y,a[x][y]); 
	dfs(x-1,y,a[x][y]); 
	dfs(x,y+1,a[x][y]); 
	dfs(x,y-1,a[x][y]);
	return; 
} 
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<=n;j++)
		   cin>>a[i][j];
	while(m--)
	{
		ll x,y;
		cin>>x>>y;
		if(jz[x][y]==0)
		{
		  dfs(x,y,'a');
      for (auto&i:b)
        jz[i.first][i.second]=ans;
      b.clear();
      ans=0;
		}
	  printf("%lld\n",jz[x][y]);
	 } 
	 return 0;
}

洪水填充板子题：bfs+区域标记

int dx[4] = { 0,0,-1,1 };
int dy[4] = { 1,-1,0,0 };

void solveA() {
	int t = 1;
	et(t) {
		int n, m; cin >> n >> m;
		vector<vector<int>>matrix(n + 1, vector<int>(n + 1));
		for (int i = 1; i <= n; i++) {
			string str; cin >> str;
			for (int j = 1; j <= n; j++) {
				matrix[i][j] = str[j - 1] + '0';
			}

		}

		int cnt = 1;
		unordered_map<int, int>hash;
		vector<vector<int>>mask(n + 1, vector<int>(n + 1,0));
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if (mask[i][j] == 0) {
					queue<pair<int,int>>q;
					int num = 1;
					q.push({ i,j });
					mask[i][j] = cnt;
					
					while (!q.empty()) {
						int top_i = q.front().first;
						int top_j = q.front().second;
						q.pop();

						for (int dic = 0; dic < 4; dic++) {
							int ni = top_i + dy[dic];
							int nj = top_j + dx[dic];

							if (ni >= 1 && ni <= n && nj >= 1 && nj <= n&&mask[ni][nj]==0) {
								if (matrix[top_i][top_j] != matrix[ni][nj]) {
									q.push({ ni,nj });
									mask[ni][nj] = cnt;
									num++;
								}
							}
						}
					}
					hash[cnt] = num;
					cnt++;
				}
			}
		}

		for (int i = 0; i < m; i++) {
			int ni, nj; cin >> ni >> nj;
			cout << hash[mask[ni][nj]] << endl;
		}
	}
}

BFS

ll n,m;
const ll Max1=1005;
char t[Max1][Max1];
ll pd[Max1][Max1];
ll dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};

void bfs(ll x,ll y)
{
  ll ans=1;
  vector<pair<ll,ll>>jyh;
  queue<pair<ll,ll>>dl;
  dl.push(make_pair(x,y));
  jyh.push_back(make_pair(x,y));
  pd[x][y]=1;
  while(!(dl.empty()))
  {
  	ll x1,y1;
  	x1=dl.front().first,y1=dl.front().second;
  	dl.pop();
  	for(int i=0;i<4;i++)
    {
      if(((x1+dx[i]>=1&&x1+dx[i]<=n)&&(y1+dy[i]>=1&&y1+dy[i]<=n))&&t[x1][y1]!=t[x1+dx[i]][y1+dy[i]]&&pd[x1+dx[i]][y1+dy[i]]==0)
      {
        ans++;
        dl.push(make_pair(x1+dx[i],y1+dy[i]));
        jyh.push_back(make_pair(x1+dx[i],y1+dy[i]));
        pd[x1+dx[i]][y1+dy[i]]=1;
      }
    }
  }
  for(auto i:jyh)
  	pd[i.first][i.second]=ans;
  return ;
}
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<=n;++j)
        cin>>t[i][j];
    while(m--)
    {
      ll x,y;
      cin>>x>>y;
      if(pd[x][y]==0)
        bfs(x,y);
      cout<<pd[x][y]<<'\n';
    }
    return 0;
} 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1010;
int n,m,ans,sum[1010100];
char mp[N][N];
bool vis[N][N];
int dx[] = {0,0,1,-1},dy[] = {1,-1,0,0};
int f[1010100];
int find(int p)
{
	if(p!=f[p]) f[p] = find(f[p]);
	return f[p];
}
void dfs(int x,int y)
{	
	if(ans == n*n) return;
	for(int i=0;i<4;i++)
	{
		int tx = x+dx[i],ty = y+dy[i];
		if(tx<=0||tx>n||ty<=0||ty>n) continue;
		if(vis[tx][ty]) continue;
		if(mp[tx][ty] == mp[x][y]) continue;
		vis[tx][ty] = 1;
		f[(tx-1)*n+ty] = find((x-1)*n+y);
		ans ++;
		dfs(tx,ty);
	}
	return;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	
	cin>>n>>m;
	for(int i = 1;i<=n;i++)
	{
		string s;
		cin>>s;
		for(int j = 0;j<n;j++)
		{
			mp[i][j+1] = s[j];
		}
	}	
	for(int i = 1;i<=n;i++)
	{
		for(int j = 1;j<=n;j++)
		{
			int idx = (i-1)*n+j;
			f[idx] = idx;
		}
	}
	for(int i = 1;i<=m;i++)
	{
		int x,y;
		ans = 1;
		cin>>x>>y;
		if(sum[find((x-1)*n+y)]) cout<<sum[find((x-1)*n+y)]<<endl;
		else
		{
			vis[x][y] = 1;
			ans = 1;
			dfs(x,y);
			sum[(x-1)*n+y] = ans;
			cout<<ans<<endl;
		}
		
	}
	return 0;
}

dfs+并查集优化（（（））））