HDU 5289 ideas + line Tree processing

Give the number of N, and K

Find the number of intervals for all satisfying conditions (maximum-minimum number <k) of the n number

Array b records start at the current position, to the right of the maximum number of conditions, the value of B array can be found by the binary + segment tree to find the maximum interval minimum value

For the second set of data

10 5

0 3 4 5 2 1 6 7 8 9

The b array is

3 7 7 7 3 1 4 3 2 1

Because the current I point takes the most right-hand value may cause the i+1 point and the subsequent fetch to the point does not satisfy the condition, all should have: b[i]<=b[i+1]+1

Get B Array

3 5 4 3 2 1 4 3 2 1

Ans=28

#include "stdio.h" #include "string.h" struct node{int l,r; __int64 Ma,mi;}    Data[400010];__int64 a[100010],b[100010];__int64 Max (__int64 A,__int64 b) {if (a<b) return B; else return A;}    __int64 Min (__int64 A,__int64 b) {if (a<b) return A; else return b;}    void build (int l,int r,int k) {int mid;    Data[k].l=l;    Data[k].r=r;        if (l==r) {DATA[K].MA=DATA[K].MI=A[DATA[K].L];    return;    } mid= (DATA[K].L+DATA[K].R)/2;    Build (L,MID,K*2);    Build (mid+1,r,k*2+1);    Data[k].ma=max (data[k*2].ma,data[k*2+1].ma); Data[k].mi=min (DATA[K*2].MI,DATA[K*2+1].MI);}    __int64 query (int l,int r,int k,int op) {int mid;        if (data[k].l==l && data[k].r==r) {if (op==0) return DATA[K].MI;    else return data[k].ma;    } mid= (DATA[K].L+DATA[K].R)/2;    if (R&LT;=DATA[K*2].R) return query (L,R,K*2,OP);    else if (L&GT;DATA[K*2].R) return query (L,R,K*2+1,OP); else {if (op==0) return Min (query (l,mid,k*2,OP), query (MID+1,R,K*2+1,OP));    else return Max (query (L,MID,K*2,OP), query (MID+1,R,K*2+1,OP));    }}int Main () {int t,n,k,i,l,r,mid;    __int64 Ans,ma,mi;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&k);        for (i=1;i<=n;i++) scanf ("%i64d", &a[i]);        Build (1,n,1);            for (i=1;i<n;i++) {l=i;r=n;            if (data[1].ma<a[i]+k && data[1].mi>a[i]-k) l=n+1;                while (l<=r) {mid= (l+r)/2;                Ma=query (l,mid,1,1);                Mi=query (l,mid,1,0);                if (ma>=a[i]+k | | mi<=a[i]-k) r=mid-1;            else l=mid+1;        } b[i]=l-i;        } b[n]=1;        for (i=n-1;i>=1;i--) if (b[i]>b[i+1]+1) b[i]=b[i+1]+1;        ans=0;        for (i=1;i<=n;i++) ans+=b[i];    printf ("%i64d\n", ans); } return 0;}

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HDU 5289 ideas + line Tree processing