Reprint Please specify the source Thank you: http://blog.csdn.net/vmurder/article/details/42886393
Exercises
Well, this picture is really not very good to do,
But we can convert it to perfection is a 1/0 sub-matrix problem.
is to reverse the point 01 of the same row parity, and then it's OK (this is obviously, need to prove the message).
And then we're asking for the maximum sub-matrix.
As for the square? To seek the sub-matrix by the way, that is the ans1,
This is obviously because we enumerate the ductility of each point, so we don't miss out on any possibility.
So let's say that if a square is not considered, its length or width must be a maximum value, then it will definitely appear in one of our calculated rectangles!
And then how to find the maximum 01 sub-matrices See previous blog http://blog.csdn.net/vmurder/article/details/42884845
Code:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define N 2020# Define INF 0x3f3f3f3fusing namespace Std;int map[n][n];int l[n],r[n],d[n];int n,m,ans1,ans2;int Main () {Freopen (" Test.in "," R ", stdin), int i,j,k,t;scanf ("%d%d ", &n,&m), for (i=1;i<=n;i++) for (j=1;j<=m;j++) {scanf ("%d " , &map[i][j]); if ((i+j&1) ==0) map[i][j]= (!map[i][j]);} D[0]=d[m+1]=0;for (i=1;i<=n;i++) {for (j=1;j<=m;j++) {l[j]=r[j]=j;if (map[i][j]) D[j]++;else d[j]=0;} for (j=1;j<=m;j++) if (D[j]) while (D[l[j]-1]>=d[j]) l[j]=l[l[j]-1];for (j=m;j>=1;j--) if (D[j]) while (d[r[j]+1 ]>=D[J]) r[j]=r[r[j]+1];for (j=1;j<=m;j++) {k=r[j]-l[j]+1;t=min (k,d[j]); Ans1=max (ans1,t*t); Ans2=max (ans2,k* D[J]);}} memset (d,0,sizeof D); for (i=1;i<=n;i++) {for (j=1;j<=m;j++) {l[j]=r[j]=j;if (!map[i][j]) D[j]++;else d[j]=0;} for (j=1;j<=m;j++) if (D[j]) while (D[l[j]-1]>=d[j]) l[j]=l[l[j]-1];for (j=m;j>=1;j--) if (D[j]) while (d[r[j]+1 ]>=D[J]) r[j]=r[r[j]+1];for (j=1;j<=m;j++) {k=r[j]-l[J]+1;t=min (K,d[j]); Ans1=max (ans1,t*t); Ans2=max (Ans2,k*d[j]);}} printf ("%d\n%d\n", ans1,ans2); return 0;}
"BZOJ1057" "ZJOI2007" checkerboard making up to 01 sub-matrices/squares