bzoj2096 pilots

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=3000200;
 9 struct que{
10     int v;//val
11     int p;//pos
12 };
13 que mx[mxn],mi[mxn];
14 int h1,h2,t1,t2;
15 int k,n;
16 int a[mxn];
17 int ans=0;
18 int main(){
19     scanf("%d%d",&k,&n);
20     int i,j;
21     for(i=1;i<=n;i++)scanf("%d",&a[i]);
22     h1=1;h2=1;t1=0;t2=0;
23     int le=0;
24     for(i=1;i<=n;i++){
25         //max
26         while(h1<=t1 && a[i]>=mx[t1].v)t1--;
27         mx[++t1].v=a[i];
28         mx[t1].p=i;
29         //min
30         while(h2<=t2 && a[i]<=mi[t2].v)t2--;
31         mi[++t2].v=a[i];
32         mi[t2].p=i;
33         
34         printf("test1 %d\n",mx[h1].v-mi[h2].v);
35         while(mx[h1].v-mi[h2].v>k){//若两队首元素差大于k，更新队头 
36             le=min(mx[h1].p,mi[h2].p)+1;
37             while(le>mx[h1].p)h1++;//队首维护 
38             while(le>mi[h2].p)h2++;
39             
40         }
41         printf("test2 %d\n",mx[h1].v-mi[h2].v);
42         ans=max(ans,i-le+1);
43     }
44     printf("%d\n",ans);
45     return 0;
46 }