Vijos 1255 mooncake Box maximum sub-matrix

Test Instructions:Link Method:Maximum sub-matrix parsing:have done so many of the largest sub-matrix, this problem casually yy came out, but I have an idea ah, if a n*m matrix, there are some bad points, non-bad points of the right value, can be negative, then the selection of the maximum and how much of the non-bad idea matrix? It seems that DP? First of all, because the bad point weight is 0, you can preprocess the sum[i,j] for the lower right corner, (I,J) for the upper left corner of the matrix weights and. We can, of course, brute-force enumeration matrices, but this is a 4-time side. Again observed n,m<=1000, so we can choose to calculate ♂ method ② to do this problem. The sum is preprocessed first, and the le,ri,h. Then enumerate all the good points and calculate the four vertex coordinates of their maximal sub-matrices, and a simple allowance is a random one. Code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 1010Using namespace Std;int n,m;int map[N][N];bool m[N][N];int sum[N][N];int le[N][N];int ri[N][N];int h[N][N];void Getsum () {for (int i=1;i<=n;i++)    {int s=0;for (int j=1;j<=m;j++)        {S+=map[i][j];Sum[i][j]=s+sum[i-1][j];        }     }}void Init () {for (int i=1;i<=n;i++)    {for (int j=1;j<=m;j++)        {if (! M[I][J])le[i][j]=0;else le[i][j]=le[i][j-1]+1;        }for (int j=m;j>=1;j--)        {if (! M[I][J])ri[i][j]=0;else ri[i][j]=ri[i][j+1]+1;        }    }for (int i=1;i<=n;i++)    {for (int j=1;j<=m;j++)        {if (M[i][j]&&m[i-1][j])            {h[i][j]=h[i-1][j]+1;le[i][j]=min (Le[i][j],le[i-1][j]);ri[i][j]=min (Ri[i][j],ri[i-1][j]);            }        }    }} int main () {scanf ("%d%d", &n,&m);for (int i=1;i<=n;i++)    {for (int j=1;j<=m;j++)        {scanf ("%d", &map[i][j]);m[i][j]=!map[i][j]?0:1;        }    }getsum ();init ();int ans=0;for (int i=1;i<=n;i++)    {for (int j=1;j<=m;j++)        {if (M[i][j])            {int ptx1,pty1,ptx2,pty2,ptx3,pty3,ptx4,pty4;Ptx1=i,pty1=j-le[i][j];Ptx2=i-h[i][j]-1,pty2=j-le[i][j];ptx3=i-h[i][j]-1,pty3=j+ri[i][j]-1;ptx4=i,pty4=j+ri[i][j]-1;Ans=max (Ans,sum[ptx4][pty4]-sum[ptx1][pty1]-sum[ptx3][pty3]+sum[ptx2][pty2]);            }        }    }printf ("%d\n", ans);}

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Vijos 1255 mooncake Box maximum sub-matrix