【BZOJ2096】[POI2010] Pilots

堆
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e6+5;
int k,n,ans = 1,a[maxn];
struct data {
    int pos,val;
    data(int a,int b):pos(a),val(b){}    
};
bool operator<(data a,data b) {return a.val < b.val;}
priority_queue<data> q1,q2;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void work(){
    int t(1);
    for(int i = 1;i <= n; ++i){
        q1.push(data(i,a[i]));
        q2.push(data(i,-a[i]));
        while(q1.top().val+q2.top().val>k){
            t++;
            while(!q1.empty() && q1.top().pos < t) q1.pop();
            while(!q2.empty() && q2.top().pos < t) q2.pop();
        }
        ans = max(ans,i-t+1);
    }
}
int main(){
//    freopen("pil.in","r",stdin);
//　  freopen("pil.out","w",stdout);
    k = read(),n = read();
    for(int i = 1;i <= n; ++i) a[i] = read();
    work();
    printf("%d",ans);
    return 0;
}

单调队列（手写）
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e6+5;
int a[maxn],q1[maxn],q2[maxn];
int k,n,ans = 1;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void work(){
    int l1(1),l2(1),r1(0),r2(0),t(1);
    for(int i = 1;i <= n; ++i){
        while(l1<=r1 && a[i] >= a[q1[r1]]) r1--;
        while(l2<=r2 && a[i] <= a[q2[r2]]) r2--;
        q1[++r1] = i,q2[++r2] = i;
        while(a[q1[l1]]-a[q2[l2]] > k)
            if(q1[l1]<q2[l2]) t = q1[l1]+1,l1++;
            else t = q2[l2]+1,l2++;
        ans = max(ans,i-t+1);
    }
}
int main(){
//    freopen("pil.in","r",stdin);
//    freopen("pil.out","w",stdout); 
    k = read(),n = read();
    for(int i = 1;i <= n; ++i) a[i] = read();
    work();
    printf("%d",ans);
    return 0;
}

双端队列
#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e6+5;
int k,n,ans=0;
int a[maxn];
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
deque<int> q1;
deque<int> q2;
void work(){
    int t;
    for(int i=1;i<=n;i++){
    while((!q1.empty())&&a[q1.front()]<=a[i]) q1.pop_back();//单减 
    while((!q2.empty())&&a[q2.front()]>=a[i]) q2.pop_back();//单增 
    q1.push_back(i); q2.push_back(i);
    while( a[q1.front()] - a[q2.front()] > k ){
        if(q2.front()>q1.front())    t=q1.front(),q1.pop_front();
        else    t=q2.front(),q2.pop_front();
        }
        ans = max(ans,i-t);
    }
}
int main(){
//    freopen("pil.in","r",stdin);
//    freopen("pil.out","w",stdout);
    k = read(),n = read();
    for(int i = 1;i <= n; ++i) a[i] = read();
    work();
    printf("%d\n",ans);
    return 0;
}