HDU3308 LCIS

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

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<=10 ⁵). 
The next line has n integers(0<=val<=10 ⁵). 
The next m lines each has an operation: 
U A B(0<=A,n , 0<=B=10 ⁵) 
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

Source

HDOJ Monthly Contest – 2010.02.06

 

 

线段树维护每个区间的三个值：从左开始数的最长单调上升序列，本区间内最长答案，以最右为结尾的最长单调上升序列。

每次合并的时候，如果左子树的右端点的值小于右子树的左端点的值，那么两段序列可以接起来……

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ls l,mid,rt<<1
#define rs mid+1,r,rt<<1|1
using namespace std;
const int mxn=100010;
int n,m;
int d[mxn];
struct node{
    int cnt;int lans,rans;
}t[mxn<<2];
void update(int p,int l,int r,int rt){
    if(l==r)return;
    int mid=(l+r)>>1;
    if(p<=mid)update(p,ls);
    else update(p,rs);
    t[rt].cnt=max(t[rt<<1].cnt,t[rt<<1|1].cnt);
    t[rt].lans=t[rt<<1].lans;t[rt].rans=t[rt<<1|1].rans;
    if(d[mid]<d[mid+1]){
        t[rt].cnt=max(t[rt].cnt,t[rt<<1].rans+t[rt<<1|1].lans);
        if(t[rt<<1].lans==mid-l+1)
            t[rt].lans=t[rt<<1].lans+t[rt<<1|1].lans;
        if(t[rt<<1|1].rans==r-mid)
            t[rt].rans=t[rt<<1|1].rans+t[rt<<1].rans;
    }
    return;
}
void Build(int l,int r,int rt){
    if(l==r){
        t[rt].cnt=t[rt].lans=t[rt].rans=1;
        return;
    }
    int mid=(l+r)>>1;
    Build(ls);
    Build(rs);
    t[rt].cnt=max(t[rt<<1].cnt,t[rt<<1|1].cnt);
    t[rt].lans=t[rt<<1].lans;
    t[rt].rans=t[rt<<1|1].rans;
    if(d[mid]<d[mid+1]){
        t[rt].cnt=max(t[rt].cnt,t[rt<<1].rans+t[rt<<1|1].lans);
        if(t[rt<<1].lans==mid-l+1)
            t[rt].lans=t[rt<<1].lans+t[rt<<1|1].lans;
        if(t[rt<<1|1].rans==r-mid)
            t[rt].rans=t[rt<<1|1].rans+t[rt<<1].rans;
    }
    return;
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l && r<=R) return t[rt].cnt;
    int mid=(l+r)>>1;
    if(R<=mid)return query(L,R,ls);
    if(mid<L)return query(L,R,rs);
    else{
        int res=max(query(L,R,ls),query(L,R,rs));
        if(d[mid]<d[mid+1]){//如果结点所存答案长于区间长度，实际答案等于区间长度，用min维护
            res=max(res,(min(t[rt<<1].rans,mid-L+1)+min(t[rt<<1|1].lans,R-mid)));
        }
        return res;
    }
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        memset(t,0,sizeof 0);
        int i,j;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)scanf("%d",&d[i]);
        Build(1,n,1);
        char op[2];int a,b;
        for(i=1;i<=m;i++){
            scanf("%s%d%d",op,&a,&b);
            if(op[0]=='U'){
                d[++a]=b;
                update(a,1,n,1);
            }
            if(op[0]=='Q'){
                
                printf("%d
",query(a+1,b+1,1,n,1));
            }
        }    
    }
    return 0;
}