[USACO12MAR]花盆Flowerpot (单调队列,二分答案)

题目链接

Solution

转化一下,就是个单调队列.

可以发现就是一段区间 \([L,R]\) 使得其高度的极差不小于 \(d\) ,同时满足 \(R-L\) 最小.

然后可以考虑二分然后再 \(O(n)\) 判断, 时间复杂度 \(O(nlogn)\) .

Code

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1000008;

ll n,d;
ll ans;
struct point{
    ll x,y;
}a[maxn];
bool jud(int len)
{
    int qmax[maxn],qmin[maxn];
    int h1,h2,t1,t2;
    h1=h2=1,t1=t2=0;
    qmax[0]=qmin[0]=0;
    for(int i=1;i<=n;++i)
    {
        while(h1<=t1 && a[i].y>a[qmax[t1]].y)--t1;
        qmax[++t1]=i;
        while(h2<=t2 && a[i].y< a[qmin[t2]].y)--t2;
        qmin[++t2]=i;
        while(h1<=t1 && a[i].x-a[qmax[h1]].x>len)++h1;
        while(h2<=t2 && a[i].x-a[qmin[h2]].x>len)++h2;
        if(a[qmax[h1]].y-a[qmin[h2]].y>=d) return 1;
    }
    return 0;
}

bool cmp(point s,point t)
{return s.x<t.x;}

int main()
{
    scanf("%lld%lld",&n,&d);
    for(int i=1;i<=n;i++)
        scanf("%lld%lld",&a[i].x,&a[i].y);
    sort(a+1,a+n+1,cmp);
    int l=1,r=maxn;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(jud(mid)){ans=mid;r=mid-1;}
        else l=mid+1;
    }
    if(ans)
    printf("%lld\n",ans);
    else puts("-1\n");
    return 0;
}