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;
}

#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（（（））））