HDOJ 题目3308 LCIS（线段树，区间查询，区间合并）

Posted 2020-09-22 llguanli

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5319    Accepted Submission(s): 2361

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

 

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ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?
a:b)
#define min(a,b) (a>b?
b:a)
struct s
{
	int lnum,rnum;
	int lmax,rmax,mmax;
}node[1000100<<2];
void init(int tr,int num)
{
	node[tr].lnum=num;
	node[tr].rnum=num;
	node[tr].lmax=1;
	node[tr].rmax=1;
	node[tr].mmax=1;
}
void pushup(int tr,int m)
{
	node[tr].lnum=node[tr<<1].lnum;
	node[tr].rnum=node[tr<<1|1].rnum;
	node[tr].lmax=node[tr<<1].lmax;
	node[tr].rmax=node[tr<<1|1].rmax;
	node[tr].mmax=max(node[tr<<1].mmax,node[tr<<1|1].mmax);
	if(node[tr<<1].rnum<node[tr<<1|1].lnum)
	{
		node[tr].mmax=max(node[tr].mmax,node[tr<<1].rmax+node[tr<<1|1].lmax);
		if(node[tr<<1].lmax==m-(m>>1))
			node[tr].lmax+=node[tr<<1|1].lmax;
		if(node[tr<<1|1].rmax==(m>>1))
			node[tr].rmax+=node[tr<<1].rmax;
	}
}	
void build(int l,int r,int tr)
{
	if(l==r)
	{
		int num;
		scanf("%d",&num);
		init(tr,num);
		return;
	}
	int mid=(l+r)>>1;
	build(l,mid,tr<<1);
	build(mid+1,r,tr<<1|1);
	pushup(tr,r-l+1);
}
void update(int pos,int num,int l,int r,int tr)
{
	if(l==r)
	{
		init(tr,num);
		return;
	}
	int mid=(l+r)>>1;
	if(pos<=mid)
	{
		update(pos,num,l,mid,tr<<1);
	}
	else
		update(pos,num,mid+1,r,tr<<1|1);
	pushup(tr,r-l+1);
}
int query(int L,int R,int l,int r,int tr)
{
	if(L<=l&&R>=r)
	{
		return node[tr].mmax;
	}
	int mid=(l+r)>>1;
	int temp1=0,temp2=0,temp3=0;
	if(L<=mid)
		temp1=query(L,R,l,mid,tr<<1);
	if(R>mid)
		temp2=query(L,R,mid+1,r,tr<<1|1);
	if(L<=mid&&R>mid&&node[tr<<1].rnum<node[tr<<1|1].lnum)
		temp3=min(mid-L+1,node[tr<<1].rmax)+min(R-mid,node[tr<<1|1].lmax);
	int ans=max(temp1,max(temp2,temp3));
	return ans;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m;
		scanf("%d%d",&n,&m);
		build(1,n,1);
		while(m--)
		{
			char s[2];
			int a,b;
			scanf("%s%d%d",s,&a,&b);
			if(s[0]==‘U‘)
			{
				update(a+1,b,1,n,1);
			}
			else
			{
				printf("%d\n",query(a+1,b+1,1,n,1));
			}
		}
	}
}