P2698 [USACO12MAR] Pots Flowerpot-- queue monotone

Record day to see (copy) solution to a problem of daily;

https://www.luogu.org/problem/P2698

We can coordinate x according to the order of increasing row number order, so we left the ordinate;

If the abscissa (l, r) interval, the maximum value minus the minimum ordinate is greater than d, then the answer can be updated;

As seen growth l, r must be incremented;

It may prove absent (l2, r2), l2> l1 and r2 <r, (maxy-miny)> d, and can affect the answers;

Because if this exists, then r2 l should be first matching objects, the answer is better;

 

This involves a sliding window method;

Ql with [] the recording position of the maximum vertical coordinate, q2 [] the recording position of the minimum vertical coordinate;

In the range of l to R & lt;

Update the queue interval range:

while(h1<=t1&&q1[h1]<l) h1++;
while(h2<=t2&&q2[h2]<l) h2++;

 

Updating r, maximum, minimum;

    while(a[q1[h1]].y-a[q2[h2]].y<d&&r<n)
        {
            ++r;
            while(a[q1[t1]].y<a[r].y&&h1<=t1) t1--;
            q1[++t1]=r;
            while(a[q2[t2]].y>a[r].y&&h2<=t2) t2--;
            q2[++t2]=r;
        }

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e6+10;
int q1[maxn],q2[maxn],h1=1,h2=1,t1,t2;
int n,d;
struct node
{
    int x,y;
}a[maxn];
bool cmp(node qw,node we)
{
    return qw.x<we.x;
}
int ans=2147483647;
int main()
{
    scanf("%d%d",&n,&d);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i].x,&a[i].y);
    }
    sort(a+1,a+n+1,cmp);
    for(int l=1,r=0;l<=n;l++)
    {
        while(h1<=t1&&q1[h1]<l) h1++;
        while(h2<=t2&&q2[h2]<l) h2++;
        while(a[q1[h1]].y-a[q2[h2]].y<d&&r<n)
        {
            ++r;
            while(a[q1[t1]].y<a[r].y&&h1<=t1) t1--;
            q1[++t1]=r;
            while(a[q2[t2]].y>a[r].y&&h2<=t2) t2--;
            q2[++t2]=r;
        }
        if(a[q1[h1]].y-a[q2[h2]].y>=d) ans=min(ans,a[r].x-a[l].x); 
    }
    if(ans==2147483647) printf("-1");
    else printf("%d",ans);
    return 0;
}