#include <stdio.h>
#include <algorithm>
#include <string.h>
#define rep(ii,a,b) for (int ii=(a);ii<=(b);ii++)
#define rek(ii,a,b) for (int ii=(a);ii>=(b);ii--)
using namespace std;
int n,m,kk;
char s[2005];
int a[2001][2001],up[2001][2001],down[2001][2001],f[2001][2001];
int ans[2001],qx[2001],qy[2001],qup[2001],qdown[2001],b[2001],c[2001];
int nowans;
bool check(int row,int length){
     if (length>min(n,m)) return false;
     int l=1,r=1,ll=1,rr=1;
     qup[l]=1;qdown[1]=1;
     rep(i,1,m) b[i]=up[row][i];
     rep(i,1,m) c[i]=down[row][i];
     if (1>=length) if (b[qup[l]]+c[qdown[ll]]-1>=length) return true;
     rep(i,2,m){
         while (b[qup[r]]>=b[i] && r>=l) r--;
         r++;qup[r]=i;
         if (i>=length && qup[l]<=i-length) l++;
         ///-------------
         while (c[qdown[rr]]>=c[i] && rr>=ll) rr--;
         rr++;qdown[rr]=i;
         if (i>=length && qdown[ll]<=i-length) ll++;
         ///-------
         if (i>=length) if (b[qup[l]]+c[qdown[ll]]-1>=length) return true;
     }
     return false;
}
int main(){
     scanf("%d%d%d",&n,&m,&kk);
     nowans=0;
     rep(i,1,n){
         scanf("%s",s);
         rep(j,0,m-1) a[i][j+1]=(s[j]=='X')?0:1;
     }
     rep(i,1,kk){
         scanf("%d%d",&qx[i],&qy[i]);
         a[qx[i]][qy[i]]=0;
     }
     rep(i,1,n)
      rep(j,1,m) nowans=max(nowans,a[i][j]);
     rep(i,1,m) up[1][i]=a[1][i];
     rep(i,2,n)
     {
         rep(j,1,m){
             up[i][j]=a[i][j]==1?up[i-1][j]+1:0;
         }
     }
     rep(i,1,m) down[n][i]=a[n][i];
     rek(i,n-1,1)
     {
         rep(j,1,m){
             down[i][j]=a[i][j]==1?down[i+1][j]+1:0;
         }
     }
     rep(i,1,n) f[i][1]=a[i][1];
     rep(j,1,m) f[1][j]=a[1][j];
     rep(i,2,n)
      rep(j,2,m){
          if (a[i][j]==0) f[i][j]=0; else f[i][j]=min(min(f[i-1][j-1],f[i-1][j]),f[i][j-1])+1;
          nowans=max(nowans,f[i][j]);
      }

     ans[kk]=nowans;
     rek(ii,kk,2){
         a[qx[ii]][qy[ii]]=1;
         up[1][qy[ii]]=a[1][qy[ii]];
         rep(i,2,n) if (a[i][qy[ii]]==0) up[i][qy[ii]]=0;else up[i][qy[ii]]=up[i-1][qy[ii]]+1;
         down[n][qy[ii]]=a[n][qy[ii]];
         rek(i,n-1,1) if (a[i][qy[ii]]==0) down[i][qy[ii]]=0;else down[i][qy[ii]]=down[i+1][qy[ii]]+1;
         int t=0;
         rek(i,12,0)
          if (check(qx[ii],t+(1<<i))) t+=(1<<i);
          nowans=max(nowans,t);
          ans[ii-1]=nowans;
     }
     rep(i,1,kk) printf("%d\n",ans[i]);
}
 

但是由於智商捉急，完全沒看懂這位大神的線段樹做法。

不過很顯然，對於影響的這一行，顯然具有可以二分答案的性質，然後我們可以用一個單調佇列來判斷是否合法。

於是該題得到解決。

程式碼：