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].
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).
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
#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\n",query(l,r,1).mm); } } return 0; }