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=;
 int n,m;
 int d[mxn];
 struct node{
     int cnt;int lans,rans;
 }t[mxn<<];
 void update(int p,int l,int r,int rt){
     if(l==r)return;
     int mid=(l+r)>>;
     if(p<=mid)update(p,ls);
     else update(p,rs);
     t[rt].cnt=max(t[rt<<].cnt,t[rt<<|].cnt);
     t[rt].lans=t[rt<<].lans;t[rt].rans=t[rt<<|].rans;
     if(d[mid]<d[mid+]){
         t[rt].cnt=max(t[rt].cnt,t[rt<<].rans+t[rt<<|].lans);
         if(t[rt<<].lans==mid-l+)
             t[rt].lans=t[rt<<].lans+t[rt<<|].lans;
         if(t[rt<<|].rans==r-mid)
             t[rt].rans=t[rt<<|].rans+t[rt<<].rans;
     }
     return;
 }
 void Build(int l,int r,int rt){
     if(l==r){
         t[rt].cnt=t[rt].lans=t[rt].rans=;
         return;
     }
     int mid=(l+r)>>;
     Build(ls);
     Build(rs);
     t[rt].cnt=max(t[rt<<].cnt,t[rt<<|].cnt);
     t[rt].lans=t[rt<<].lans;
     t[rt].rans=t[rt<<|].rans;
     if(d[mid]<d[mid+]){
         t[rt].cnt=max(t[rt].cnt,t[rt<<].rans+t[rt<<|].lans);
         if(t[rt<<].lans==mid-l+)
             t[rt].lans=t[rt<<].lans+t[rt<<|].lans;
         if(t[rt<<|].rans==r-mid)
             t[rt].rans=t[rt<<|].rans+t[rt<<].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)>>;
     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+]){//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<<].rans,mid-L+)+min(t[rt<<|].lans,R-mid)));
         }
         return res;
     }
 }
 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         memset(t,,sizeof );
         int i,j;
         scanf("%d%d",&n,&m);
         for(i=;i<=n;i++)scanf("%d",&d[i]);
         Build(,n,);
         char op[];int a,b;
         for(i=;i<=m;i++){
             scanf("%s%d%d",op,&a,&b);
             if(op[]=='U'){
                 d[++a]=b;
                 update(a,,n,);
             }
             if(op[]=='Q'){

                 printf("%d\n",query(a+,b+,,n,));
             }
         }
     }
     return ;
 }