hdu 3308 线段树 区间合并+单点更新+区间查询

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6592    Accepted Submission(s): 2866

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

 

Author

shǎ崽

 

Source

HDOJ Monthly Contest – 2010.02.06

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=1e5+10,M=4e6+10,inf=1e9+10;
struct is
{
    int l,r;
    int lm,mm,rm;
}tree[N<<2];
int val[N];
void buildtree(int l,int r,int pos)
{
    tree[pos].l=l;
    tree[pos].r=r;
    if(l==r)
    {
        tree[pos].lm=tree[pos].rm=tree[pos].mm=1;
        return;
    }
    int mid=(l+r)>>1;
    buildtree(l,mid,pos<<1);
    buildtree(mid+1,r,pos<<1|1);
    tree[pos].lm=tree[pos<<1].lm;
    tree[pos].rm=tree[pos<<1|1].rm;
    if(val[tree[pos<<1].r]<val[tree[pos<<1|1].l])
    tree[pos].mm=max(tree[pos<<1].mm,max(tree[pos<<1|1].mm,tree[pos<<1].rm+tree[pos<<1|1].lm));
    else
    tree[pos].mm=max(tree[pos<<1].mm,tree[pos<<1|1].mm);
    if(val[tree[pos<<1].r]<val[tree[pos<<1|1].l]&&tree[pos<<1].lm==tree[pos<<1].r-tree[pos<<1].l+1)
    tree[pos].lm+=tree[pos<<1|1].lm;
    if(val[tree[pos<<1].r]<val[tree[pos<<1|1].l]&&tree[pos<<1|1].rm==tree[pos<<1|1].r-tree[pos<<1|1].l+1)
    tree[pos].rm+=tree[pos<<1].rm;
}
void update(int point,int pos)
{
    if(tree[pos].l==tree[pos].r)
    return;
    int mid=(tree[pos].r+tree[pos].l)>>1;
    if(point>mid)
    update(point,pos<<1|1);
    else
    update(point,pos<<1);
    tree[pos].lm=tree[pos<<1].lm;
    tree[pos].rm=tree[pos<<1|1].rm;
    if(val[tree[pos<<1].r]<val[tree[pos<<1|1].l])
    tree[pos].mm=max(tree[pos<<1].mm,max(tree[pos<<1|1].mm,tree[pos<<1].rm+tree[pos<<1|1].lm));
    else
    tree[pos].mm=max(tree[pos<<1].mm,tree[pos<<1|1].mm);
    if(val[tree[pos<<1].r]<val[tree[pos<<1|1].l]&&tree[pos<<1].lm==tree[pos<<1].r-tree[pos<<1].l+1)
    tree[pos].lm+=tree[pos<<1|1].lm;
    if(val[tree[pos<<1].r]<val[tree[pos<<1|1].l]&&tree[pos<<1|1].rm==tree[pos<<1|1].r-tree[pos<<1|1].l+1)
    tree[pos].rm+=tree[pos<<1].rm;
}
is query(int L,int R,int pos)
{
    if(tree[pos].l==L&&tree[pos].r==R)
    return tree[pos];
    int mid=(tree[pos].l+tree[pos].r)>>1;
    if(mid<L)
    return query(L,R,pos<<1|1);
    else if(mid>=R)
    return query(L,R,pos<<1);
    else
    {
        is a=query(L,mid,pos<<1);
        is b=query(mid+1,R,pos<<1|1);
        is ans;
        ans.l=a.l,ans.r=b.r;
        ans.lm=a.lm;
        ans.rm=b.rm;
        if(val[a.r]<val[b.l])
        ans.mm=max(a.mm,max(b.mm,a.rm+b.lm));
        else
        ans.mm=max(a.mm,b.mm);
        if(val[a.r]<val[b.l]&&a.lm==a.r-a.l+1)
        ans.lm+=b.lm;
        if(val[a.r]<val[b.l]&&b.rm==b.r-b.l+1)
        ans.rm+=a.rm;
        return ans;
    }
}
char a[10];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        scanf("%d",&val[i]);
        buildtree(1,n,1);
        while(m--)
        {
            int l,r;
            scanf("%s%d%d",a,&l,&r);
            l++;
            if(a[0]=='U')
            val[l]=r,update(l,1);
            else
            r++,printf("%d
",query(l,r,1).mm);
            }
    }
    return 0;
}