【2019多校】Make Rounddog Happy（分治）

Rounddog always has an array a1,a2,⋯,ana1,a2,⋯,an in his right pocket, satisfying 1≤ai≤n1≤ai≤n .

A subarray is a non-empty subsegment of the original array. Rounddog defines a good subarray as a subsegment al,al+1,⋯,aral,al+1,⋯,ar that all elements in it are different and max(al,al+1,…,ar)−(r−l+1)≤kmax(al,al+1,…,ar)−(r−l+1)≤k .

Rounddog is not happy today. As his best friend, you want to find all good subarrays of aa to make him happy. In this case, please calculate the total number of good subarrays of aa .

InputThe input contains several test cases, and the first line contains a single integer T (1≤T≤20)T (1≤T≤20) , the number of test cases.

The first line of each test case contains two integers n (1≤n≤300000)n (1≤n≤300000) and k (1≤k≤300000)k (1≤k≤300000) .

The second line contains nn integers, the ii -th of which is ai (1≤ai≤n)ai (1≤ai≤n) .

It is guaranteed that the sum of nn over all test cases never exceeds 10000001000000 .
OutputOne integer for each test case, representing the number of subarrays Rounddog likes.
Sample Input

2
5 3
2 3 2 2 5
10 4
1 5 4 3 6 2 10 8 4 5

Sample Output

7
31
SOLUTION:
很容易想到考虑每一个数作为最大值的情况，然后处理出他所能想两边扩展的区间
但是向两边的很好搞，但是必须跨区间，然后就自闭了
后来知道分治可以很好的搞这个问题
题解：https://www.cnblogs.com/Dillonh/p/11390927.html
启发式的所有东西都是玄学的复杂度。。。。。qwq

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=1000010;
int a[maxn],st[maxn][21],lg[maxn],K,N; ll ans;
int A[maxn],B[maxn],vis[maxn];
int get(int L,int R)
{
    int k=lg[R-L+1];
    return a[st[L][k]]>=a[st[R-(1<<k)+1][k]]?st[L][k]:st[R-(1<<k)+1][k];
}
void solve(int L,int R)
{
    if(L>R) return ;
    int pos=get(L,R);
    if(pos-L<R-pos){
        rep(i,L,pos){
            int t=a[pos]-K,lR=i+t-1;
            int fcy=min(R,B[i]);
            lR=max(lR,pos);
            if(lR>fcy) continue;
            ans+=fcy-lR+1;
        }
    }
    else {
        rep(i,pos,R){
            int t=a[pos]-K,rL=i-t+1;
            int fcy=max(L,A[i]);
            rL=min(rL,pos);
            if(rL<fcy) continue;
            ans+=rL-fcy+1;
        }
    }
    solve(L,pos-1); solve(pos+1,R);
}
int main()
{
    int T;
    lg[0]=-1; rep(i,1,maxn-1) lg[i]=lg[i>>1]+1;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&N,&K); ans=0;
        rep(i,1,N) scanf("%d",&a[i]);
        rep(i,1,N) st[i][0]=i;
        rep(i,1,20) {
            rep(j,1,N+1-(1<<i))
             st[j][i]=a[st[j][i-1]]>=a[st[j+(1<<(i-1))][i-1]]?st[j][i-1]:st[j+(1<<(i-1))][i-1];
        }
    
        rep(i,1,N) vis[i]=0; A[1]=1; vis[a[1]]=1;
        rep(i,2,N){
             if(vis[a[i]])  A[i]=max(A[i-1],vis[a[i]]+1);
             else A[i]=A[i-1];
             vis[a[i]]=i;
        }
        rep(i,1,N) vis[i]=0; B[N]=N; vis[a[N]]=N;
        for(int i=N-1;i>=1;i--){
            if(vis[a[i]]) B[i]=min(B[i+1],vis[a[i]]-1);
            else B[i]=B[i+1];
            vis[a[i]]=i;
        }
        solve(1,N);
        printf("%lld
",ans);
    }
    return 0;
}