[P2216] [HAOI2007] ideal square "monotonous queue"

Thinking: each row and column in a monotonous maintenance queue.

Specific method: a first value for each row of a maintenance queue monotonous, and A [] [] maximum value for each interval, respectively the minimum presence of X-[] [] and X [] [] in.

Then the X-[] [] and X [] [] are stored respectively maximum and minimum values within the rectangle of 1 × n. X-[i] [J ] of the memory i of row j ~ j + n-1 rectangular column the maximum value. Similarly, X [i] [J] storage of the i first row j ~ j + n-1 rectangular column minimum.

Then again these values on each column to maintain two arrays, the X-[] [] is the maximum value in each interval with the Y [] [] maintenance, the X [] [] in each section the minimum value Y [] [] maintenance. Then Y [i] [j] stored X-[] [] the first i ~ i + n-1 oblong maximum row j-th column. Similarly y [i] [j] stored X [] [] the first i ~ i + n-1 rectangular row j-th column of the minimum value.

Therefore, Y [i] [j] real storage as in a [i ~ i + n- 1] [j ~ j + n-1] is the largest, that is i, j is the upper left corner, the side length n the maximum value of the square. Similarly, Y [I] [J] stored i.e. i, j is the top left side length n minimum value of the square.

Simulation process shown below:

 Code

#include <bits/stdc++.h>
using namespace std;

int n,m,k,front,FRONT,back,BACK,ans;
int a[1001][1001],q[1001],Q[1001];
int x[1001][1001],X[1001][1001];
int y[1001][1001],Y[1001][1001];

int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for (int I=1;I<=n;I++)
        for (int i=1;i<=m;i++)
            scanf("%d",&a[I][i]);
    for (int I=1;I<=n;I++)
        {
            FRONT=BACK=front=back=Q[1]=q[1]=1;
            for (int i=2;i<=m;i++)
                {
                    while (a[I][i]>=a[I][Q[BACK]]&&FRONT<=BACK) BACK--;
                    while (a[I][i]<=a[I][q[back]]&&front<=back) back--;
                    BACK++;back++;Q[BACK]=i;q[back]=i;
                    while (i-Q[FRONT]>=k) FRONT++;
                    while (i-q[front]>=k) front++;
                    if (i>=k) X[I][i-k+1]=a[I][Q[FRONT]],x[I][i-k+1]=a[I][q[front]];
                }
        }
    for (int I=1;I<=m-k+1;I++)
        {
            FRONT=BACK=front=back=Q[1]=q[1]=1;
            for (int i=2;i<=n;i++)
                {
                    while (X[i][I]>=X[Q[BACK]][I]&&FRONT<=BACK) BACK--;
                    while (x[i][I]<=x[q[back]][I]&&front<=back) back--;
                    BACK++;back++;Q[BACK]=i;q[back]=i;
                    while (i-Q[FRONT]>=k) FRONT++;
                    while (i-q[front]>=k) front++;
                    if (i>=k) Y[i-k+1][I]=X[Q[FRONT]][I],y[i-k+1][I]=x[q[front]][I];
                }
        }
    ans=0x3f3f3f3f;
    for (int I=1;I<=n-k+1;I++)
        for (int i=1;i<=m-k+1;i++)
            ans=min(ans,Y[I][i]-y[I][i]);
    printf("%d\n",ans);
    return 0;
}

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