HDU 3308 LCIS

Problem Description
Given n integers.
You have two operations:
U A B: replace the Ath number by B. (index counting from 0)
Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].

Input
T in the first line, indicating the case number.
Each case starts with two integers n , m(0 < n,m<=105).
The next line has n integers(0<=val<=105).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=105)
OR
Q A B(0<=A<=B< n).

Output
For each Q, output the answer.

Sample Input
1
10 10
7 7 3 3 5 9 9 8 1 8
Q 6 6
U 3 4
Q 0 1
Q 0 5
Q 4 7
Q 3 5
Q 0 2
Q 4 6
U 6 10
Q 0 9

Sample Output
1
1
4
2
3
1
2
5

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这题。。说多了都是泪。
算法好想，没有坑点，异常复杂。
容我整理思路，来日更博。(删除)
上神犇博客这里

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#include<iostream>#include<cstdio>using namespace std;const int maxn=100005;int t,n,m,data[maxn];struct node{    int l,r,ln,rn,lm,rm,mx;}a[4*maxn];void update(int num){    a[num].ln=a[2*num].ln;    a[num].rn=a[2*num+1].rn;    a[num].lm=a[2*num].lm;    a[num].rm=a[2*num+1].rm;    a[num].mx=max(a[2*num].mx,a[2*num+1].mx);    if(a[2*num].rn<a[2*num+1].ln)    {        if(a[2*num].lm==a[2*num].r-a[2*num].l+1)            a[num].lm+=a[2*num+1].lm;        if(a[2*num+1].rm==a[2*num+1].r-a[2*num+1].l+1)            a[num].rm+=a[2*num].rm;        a[num].mx=max(a[num].mx,a[2*num].rm+a[2*num+1].lm);    }}void build(int num,int l,int r){    a[num].l=l;    a[num].r=r;    if(l==r)    {        a[num].lm=a[num].rm=a[num].mx=1;        a[num].ln=a[num].rn=data[l];        return ;    }    int mid=l+r>>1;    build(2*num,l,mid);    build(2*num+1,mid+1,r);    update(num);}void chg(int num,int p,int s){    if(a[num].l==a[num].r)    {        a[num].ln=a[num].rn=s;        return ;    }    int mid=a[num].l+a[num].r>>1;    if(p<=mid)        chg(2*num,p,s);    else        chg(2*num+1,p,s);    update(num);}int query(int num,int l,int r){    if(l<=a[num].l&&r>=a[num].r)        return a[num].mx;    int ans=0,mid=a[num].l+a[num].r>>1;    if(l<=mid)        ans=max(ans,query(2*num,l,r));    if(r>=mid+1)        ans=max(ans,query(2*num+1,l,r));    if(a[2*num].rn<a[2*num+1].ln)        ans=max(ans,min(mid-l+1,a[2*num].rm)+min(r-mid,a[2*num+1].lm));    return ans;}int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)            scanf("%d",&data[i]);        build(1,1,n);        char ch[11];        int l,r;        while(m--)        {            scanf("%s%d%d",ch,&l,&r);            if(ch[0]=='Q')                printf("%d\n",query(1,l+1,r+1));            else                chg(1,l+1,r);        }    }    return 0;}