[HDU 3308] LCIS

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4437    Accepted Submission(s): 2006

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

 

线段树区间合并

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 100005

struct node
{
    int l,r,m;
    int ln,rn;
    int lsum,rsum,msum;
    int mid()
    {
        return (l+r)>>1;
    }
}t[N<<2];
int a[N];

void pushup(int rt)
{
    t[rt].lsum=t[rt<<1].lsum;
    t[rt].rsum=t[rt<<1|1].rsum;
    t[rt].ln=t[rt<<1].ln;
    t[rt].rn=t[rt<<1|1].rn;
    t[rt].msum=max(t[rt<<1].msum,t[rt<<1|1].msum);
    int ll=t[rt<<1].r-t[rt<<1].l+1;
    int rr=t[rt<<1|1].r-t[rt<<1|1].l+1;

    if(t[rt<<1].rn<t[rt<<1|1].ln) 
    {
        if(t[rt<<1].lsum==ll)
            t[rt].lsum+=t[rt<<1|1].lsum;
        if(t[rt<<1|1].rsum==rr)
            t[rt].rsum+=t[rt<<1].rsum;
        t[rt].msum=max(t[rt].msum,t[rt<<1].rsum+t[rt<<1|1].lsum);
    }
}
void build(int l,int r,int rt)
{
    t[rt].l=l;
    t[rt].r=r;
    if(l==r)
    {
        t[rt].ln=t[rt].rn=a[l];
        t[rt].lsum=t[rt].rsum=t[rt].msum=1;
        return;
    }
    int m=t[rt].mid();
    build(l,m,rt<<1);
    build(m+1,r,rt<<1|1);
    pushup(rt);
}
void update(int rt,int pos,int c)
{
    if(t[rt].l==t[rt].r)
    {
        t[rt].ln=t[rt].rn=c;
        return;
    }
    int m=t[rt].mid();
    if(pos<=m) update(rt<<1,pos,c);
    else update(rt<<1|1,pos,c);
    pushup(rt);
}
int query(int rt,int L,int R)
{
    if(t[rt].l==L && t[rt].r==R)
    {
        return t[rt].msum;
    }
    int m=t[rt].mid();
    if(R<=m) return query(rt<<1,L,R);
    else if(L>m) return query(rt<<1|1,L,R);
    else
    {
        int sum=0,m=t[rt].mid();
        int ltmp=query(rt<<1,L,m);
        int rtmp=query(rt<<1|1,m+1,R);
        if(t[rt<<1].rn<t[rt<<1|1].ln)       //注意：取最大值时要保证是在给定的区间[L,R]里面,左边，取min(全长，rsum），右边取min(全长,lsum)
        {
            sum=min(m-L+1,t[rt<<1].rsum)+min(R-m,t[rt<<1|1].lsum);
        }
        return max(max(ltmp,rtmp),sum);
    }
}
int main()
{    
    int T,n,m,i;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        build(0,n-1,1);
        while(m--)
        {
            char op;
            int a,b;
            scanf(" %c%d%d",&op,&a,&b);
            if(op=='U') update(1,a,b);
            else if(op=='Q') printf("%d
",query(1,a,b));
        }
    }
    return 0;
}