bzoj1057: [ZJOI2007] Board making

Dp. Square Chessboard Needless to say, each point of a rectangular chessboard maintains the maximum distance V that can be reached first. The DP then finds the maximum distance l,r with its own v-min to the left to the right.

Because the width of the maximum rectangular checkerboard is certainly equal to the V of a point, we have also found the longest distance (r-l+1) corresponding to each V, so correctness is guaranteed.

There is also a little trick, which puts the horizontal ordinate and the odd points to vary or. You can turn the original problem into the largest 01 matrix.

#include <cstdio>#include<algorithm>#include<cstring>using namespacestd;Const intMAXN = -+Ten;intA[MAXN][MAXN],F[MAXN][MAXN],V[MAXN][MAXN],L[MAXN][MAXN],R[MAXN][MAXN];intn,m,res1=0, res2=0;intMain () {scanf ("%d%d",&n,&m);  for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++) {scanf ("%d",&A[i][j]); if((i+j)%2) a[i][j]^=1; //printf ("a[%d][%d] =%d\n", i,j,a[i][j]);    }     for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++) {        if(i==1&&j==1) f[i][j]=1; Else if(a[i][j]!=a[i][j-1] || A[I][J]! = a[i-1][J]) f[i][j]=1; ElseF[i][j] = min (f[i-1][j-1],min (f[i-1][j],f[i][j-1]))+1; Res1=Max (res1,f[i][j]); //printf ("f[%d][%d] =%d\n", i,j,f[i][j]);} printf ("%d\n", res1*res1);  for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++) {        if(i==1) v[i][j]=1; Else if(a[i-1][J]!=A[I][J]) v[i][j]=1; Elsev[i][j]=v[i-1][j]+1; }     for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++) {        if(j==1) l[i][j]=1; Else if(v[i][j]>v[i][j-1] || a[i][j]!=a[i][j-1]) l[i][j]=J; Else{L[i][j]=l[i][j-1];  while(l[i][j]>1&& v[i][j]<=v[i][l[i][j]-1] && a[i][l[i][j]-1]==A[I][J]) l[i][j]=l[i][l[i][j]-1]; }    }     for(intI=1; i<=n;i++)     for(intj=m;j>=1; j--) {        if(j==m) r[i][j]=m; Else if(v[i][j]>v[i][j+1] || A[I][J]! = a[i][j+1]) r[i][j]=J; Else{R[i][j]= r[i][j+1];  while(R[i][j]<n && v[i][j]<=v[i][r[i][j]+1] && a[i][r[i][j]+1]==A[I][J]) r[i][j]=r[i][r[i][j]+1]; }    }     for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++) Res2=max (res2,v[i][j]* (r[i][j]-l[i][j]+1)); printf ("%d\n", Res2); return 0;}

bzoj1057: [ZJOI2007] Board making