CODEFORCES480E Parking IoT

Title: To a lattice, where there is no point, the operation is to remove a point, and ask the current lattice of the largest square

If not modified, the bare DP

Plus the changes, you can consider the time reversal, so the answer is incremental

can use and check the connectivity of the maintenance point, o^2

#include <bits/stdc++.h>using namespacestd;#defineMAXN 2010inlinevoidMIN (int&a,intb) {if(a>b) a=b;} InlinevoidMAX (int&a,intb) {if(a<b) a=b;}intN,m,q,sz;intF[MAXN][MAXN],U[MAXN][MAXN],LG[MAXN],RG[MAXN],QX[MAXN],QY[MAXN],ANS[MAXN];CharS[MAXN][MAXN];structline{intF[MAXN]; intFindintx) {returnf[x]==x?x:f[x]=find (F[x]);}} L[MAXN],R[MAXN];voidDelintXinty) {L[x].f[y]=l[x].find (y1); R[x].f[y]=r[x].find (y+1);}intMain () {scanf ("%d%d%d",&n,&m,&q);  for(intI=1; i<=n;++i) for(intj=1; j<=m+1;++j) {L[i].f[j]=J; R[I].F[J]=J; U[I][J]=i; }     for(intI=1; i<=n;++i) scanf ("%s", s[i]+1);  for(intI=1; i<=q;++i) scanf ("%d%d", &qx[i],&qy[i]), s[qx[i]][qy[i]]='X';  for(intI=1; i<=n;++i) for(intj=1; j<=m;++j)if(s[i][j]=='.') del (I,J);  for(intI=1; i<=n;++i) for(intj=1; j<=m;++j)if(s[i][j]=='.') {U[i][j]=u[i-1][j]; F[I][J]=min (f[i-1][j-1]+1, Min (i-u[i][j],j-L[i].find (j)));            MAX (Sz,f[i][j]); }     for(intv=q;v;--v) {Ans[v]=sz;        Del (Qx[v],qy[v]);  for(intI=1; i<=n;++i) {Lg[i]=qy[v]-L[i].find (Qy[v]); Rg[i]=r[i].find (Qy[v])-Qy[v]; }         for(inti=qx[v]-1; i>=1; i) MIN (lg[i],lg[i+1]), MIN (rg[i],rg[i+1]);  for(inti=qx[v]+1; i<=n;++i) MIN (lg[i],lg[i-1]), MIN (rg[i],rg[i-1]);  for(intI=1; i<=qx[v];++i) while(I+sz<=n&&min (Rg[i],rg[i+sz]) +min (Lg[i],lg[i+sz])-1>sz)++sz; }     for(intI=1; i<=q;++i) printf ("%d\n", Ans[i]); return 0;}

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CODEFORCES480E Parking IoT