Problem: JOIOI Kingdom (half)

topic

Portal

Thinking

If we start thinking from positive,

DP is to start thinking like a chicken, like the author of this konjac

You will find that the value of each DP needs to have four auxiliary array to transfer

But are also affect the current value of the subsequent DP DP value

But think for a train of thought,

It is easy to find the answer is monotonic

That dichotomy

The final shape of a ladder must be

Then you find the minimum value in the trapezoid, then half of the answer to find the maximum

Time is on the complexity verifier \ (O (n * m) \)

Plus binary complexity \ (O (log_ {maxa} ) \)

If 1S, there are still regular card

But the topic of human goodness opened a 4S

Code

#include<iostream>
#include<climits>
#include<cstring>
using namespace std;
int n,m;
int a[2005][2005];
int b[2005][2005];
int c[2005][2005];
int d[2005][2005];
int h[2005];
int minn=INT_MAX;
int maxx=INT_MIN;
int l=INT_MAX;
int r=INT_MIN;
int mid;
bool dfs(int x,int t[2005][2005])
{
    memset(h,0,sizeof(h));
    h[n+1]=m;
    for(int i=n;i>=1;i--)
    {
        while(h[i]<h[i+1]&&t[i][h[i]+1]-minn<=x)
            h[i]++;
    }
    int minnh=INT_MAX;
    for(int i=1;i<=n;i++)
        for(int j=h[i]+1;j<=m;j++)
            minnh=min(minnh,t[i][j]);
    if(maxx-minnh<=x)
        return 1;
    return 0;   
}
bool pd(int cnt)
{
    if(dfs(cnt,a)||dfs(cnt,b)||dfs(cnt,c)||dfs(cnt,d))
        return 1;
    return 0;
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>a[i][j];
            d[i][j]=c[i][j]=b[i][j]=a[i][j];
            minn=min(minn,a[i][j]);
            maxx=max(maxx,a[i][j]);
        }
    }
    for(int i=1;i<=n/2;i++)
        swap(b[i],b[n-i+1]);
    for(int i=1;i<=n/2;i++)
        swap(c[i],c[n-i+1]);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m/2;j++)
        {
            swap(c[i][j],c[i][m-j+1]);
            swap(d[i][j],d[i][m-j+1]);
        }
    }
    l=0;
    r=maxx-minn;
    while(l<r)
    {
        mid=(1ll*l+r)>>1;
        if(pd(mid))
            r=mid;
        else
            l=mid+1;
    }
    cout<<l;
    return 0;
}