I. Yukino With Subinterval

題目鏈接：

Problem Descripe

Yukino has an array \(a_1, a_2 \cdots a_n\). As a tsundere girl, Yukino is fond of studying subinterval.

Today, she gives you four integers $l, r, x, y $, and she is looking for how many different subintervals \([L, R]\) are in the interval \([l, r]\)that meet the following restraints:

\(a_L =a_{L+1} =\cdots=a_R\), and for any $ i\in [L,R], x \le a_i \le y$.

The length of such a subinterval should be maximum under the first restraint.

Note that two subintervals \([L_1,R_1] , [L_2,R_2]\) are different if and only if at least one of the following formulas is true:

\(L1 \cancel= L2\)

\(R1 \cancel= R2\)

Yukino, at the same time, likes making tricks. She will choose two integers \(pos,v\), and she will change \(a_{pos}\) to \(v\).

Now, you need to handle the following types of queries:

• \(1 \ pos \ v\) : change \(a_{pos}\) to $v $
• \(2\) \(l \ r \ x \ y\): print the number of legal subintervals in the interval \([l, r]\)

Input

The first line of the input contains two integers \(n, m (1 \le n, m \le 2 \times 10^5)\)– the numbers of the array and the numbers of queries respectively.

The second line of the input contains nnn integers \(a_i (1 \le a_i \le n)\).

For the next mmm line, each containing a query in one of the following queries:

• \(1\) \(pos\) \(v \ (1 \le pos, v \le n)\): change \(a_{pos}\) to \(v\)
• \(2 \ l \ r \ x \ y (1 \le l \le r \le n) (1 \le x \le y \le n)\): print the number of legal subintervals in the interval \([l,r]\)

Output

For each query of the second type, you should output the number of legal subintervals in the interval \([l, r]\).

樣例輸入

6 3
3 3 1 5 6 5
2 2 3 4 5
1 3 2
2 1 6 1 5

樣例輸出

0
4

樣例解釋

For the first operations, there are \(3\) different subintervals \(([2, 2],[3, 3],[2,3])\)in the interval \([2, 3]\), but none of them meets all the restraints.

For the third operations, the legal subintervals in interval \([1, 6]\) are: \([1, 2], [3, 3], [4, 4], [6, 6]\)

Notes that although subintervals \([1,1]\) and \([2,2]\) also meet the first restraint, we can extend them to subinterval \([1, 2]\). So the length of them is not long enough, which against the second one.

題意

給你一個序列，提供兩種操作

• \(1\) \(pos\) \(v \ (1 \le pos, v \le n)\): 將 \(a_{pos}\) 改為 \(v\)
• \(2 \ l \ r \ x \ y (1 \le l \le r \le n) (1 \le x \le y \le n)\): 輸出\([l,r]\) 中權(quán)值\(\in [x,y]\) 的個數(shù)。特別注意一段連續(xù)相同的數(shù)只算一次

題解

樹套樹\(n\)年前打的，早就忘了，于是直接跳過，其實這就是一道可修改區(qū)間第k大模板題吧，如果不會的可以去luogu學(xué)習(xí)一下。

模板傳送門：https://www.luogu.org/problem/P3380

這題唯一要解決的就是怎么處理連續(xù)段只算一次的問題了。我是樹狀數(shù)組套線段樹，于是如果\(a[i]=a[i-1]\)那么就不處理。

還有幾個需要注意的地方

如果改變了\(a[i]\)的值，記得更改\(a[i-1]\)和\(a[i+1]\)

對于區(qū)間\([l,r]\),記得特判\(a[l]\),可能\(a[l]=a[l-1]\)，但是這時也要算

代碼

#include<bits/stdc++.h> using namespace std; #define ll long long #define INF 0x7f7f7f7f #define N 200050 template<typename T>void read(T&x) {ll k=0; char c=getchar();x=0;while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();if (c==EOF)exit(0);while(isdigit(c))x=x*10+c-'0',c=getchar();x=k?-x:x; } void read_char(char &c) {while(!isalpha(c=getchar())&&c!=EOF);} int n,m,treeNode; int a[N],ql[20],qr[20]; struct Tree{int ls,rs,sum;}tr[N*150]; void update(int&x,int p,int tt,int l,int r) {if (x==0)x=++treeNode;tr[x].sum+=tt;if (l==r)return;int mid=(l+r)>>1;if (p<=mid)update(tr[x].ls,p,tt,l,mid);else update(tr[x].rs,p,tt,mid+1,r); } void change(int x,int p,int tt) {while(x<=n)update(x,p,tt,1,n+1),x+=x&-x;} void getRt(int l,int r) {ql[0]=qr[0]=0;while(l)ql[++ql[0]]=l,l-=l&-l;while(r)qr[++qr[0]]=r,r-=r&-r; } int getSum() {int ans=0;for(int i=1;i<=ql[0];i++)ans-=tr[tr[ql[i]].ls].sum;for(int i=1;i<=qr[0];i++)ans+=tr[tr[qr[i]].ls].sum;return ans; } void move_L() {for(int i=1;i<=ql[0];i++)ql[i]=tr[ql[i]].ls;for(int i=1;i<=qr[0];i++)qr[i]=tr[qr[i]].ls; } void move_R() {for(int i=1;i<=ql[0];i++)ql[i]=tr[ql[i]].rs;for(int i=1;i<=qr[0];i++)qr[i]=tr[qr[i]].rs; } int _Rank(int p,int l,int r) {if (l==r)return 0;int mid=(l+r)>>1,tp=getSum();if (p<mid){move_L();return _Rank(p,l,mid);}move_R(); return tp+_Rank(p,mid+1,r); } int Rank(int l,int r,int k) {getRt(l-1,r);return _Rank(k-1,1,n+1); } void work() {int id,pos,v,l,r,x,y;read(n); read(m);treeNode=n;for(int i=1;i<=n;i++)read(a[i]);for(int i=1;i<=n;i++)if (a[i]!=a[i-1])change(i,a[i],1);for(int i=1;i<=m;i++){read(id);if (id==1){read(pos); read(v);if (a[pos]!=a[pos-1])change(pos,a[pos],-1);if (v!=a[pos-1])change(pos,v,1);if (a[pos]==a[pos+1])change(pos+1,a[pos+1],1);if (v==a[pos+1])change(pos+1,a[pos+1],-1);a[pos]=v;}if (id==2){read(l); read(r); read(x); read(y);int ans=-Rank(l,r,x)+Rank(l,r,y+1);if (a[l]==a[l-1]&&x<=a[l]&&a[l]<=y)ans++;printf("%d\n",ans);}} } int main() { #ifndef ONLINE_JUDGEfreopen("aa.in","r",stdin); #endifwork(); }

轉(zhuǎn)載于:https://www.cnblogs.com/mmmqqdd/p/11508864.html

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