LCIS HDU - 3308（线段树）

版权声明： https://blog.csdn.net/sgh666666/article/details/83624413

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.

线段树代码风格换成结构体了。来一道动态最大连续上升子序列。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<string>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define oo cout<<"!!!"<<endl;
typedef long long ll;
typedef unsigned long long ull;
#define ms(s) memset(s, 0, sizeof(s))
const int inf = 0x3f3f3f3f;
//head

const int maxn = 1e6+11;
int a[maxn];
struct node
{
	int l,r,dat,dl,dr,len;
}t[maxn];

void pushup(int i)
{
	int m = (t[i].l + t[i].r)>>1;
	if(a[m] < a[m+1])
	{
		t[i].dl = t[i<<1].dl == t[i<<1].len? (t[i<<1].dl + t[i<<1|1].dl) : t[i<<1].dl;
		t[i].dr = t[i<<1|1].dr == t[i<<1|1].len? (t[i<<1|1].dr + t[i<<1].dr) : t[i<<1|1].dr;
		t[i].dat = max(t[i<<1].dat,t[i<<1|1].dat);
		t[i].dat = max(t[i].dat,t[i<<1].dr + t[i<<1|1].dl);	
	}
	else
	{
		t[i].dl = t[i<<1].dl;
		t[i].dr = t[i<<1|1].dr;
		t[i].dat = max(t[i<<1].dat,t[i<<1|1].dat);
	}
}

void build(int i,int l,int r)
{
	t[i].l = l,t[i].r = r;
	t[i].len = r - l +1;
	if(l == r){t[i].dl = t[i].dr = t[i].dat = t[i].len = 1;return;}
	int  m = (l+r)>>1;
	build(i<<1,l,m);
	build(i<<1|1,m+1,r);
	pushup(i);
}


void update(int p,int v,int i)
{
	if(t[i].l == t[i].r)
	{
		return;
	}
	int  mid = (t[i].l + t[i].r)>>1;
	if(p<=mid)update(p,v,i<<1);
	else update(p,v,i<<1|1);
	pushup(i);
}

int query(int ql,int qr,int i)
{
	if(ql <= t[i].l && qr >= t[i].r)
	{
		return t[i].dat;
	}

	int mid = (t[i].l + t[i].r)>>1;
	if(qr<=mid)return query(ql,qr,i<<1);
	if(ql>mid)return query(ql,qr,i<<1|1);

	int len = query(ql,qr,i<<1);
	int ren = query(ql,qr,i<<1|1);
	int ans = max(len,ren);
	if(a[mid] < a[mid+1])
	{
		int tmp = min(t[i<<1].dr,mid-ql+1) + min(t[i<<1|1].dl,qr-mid);
		ans = max(ans,tmp);
	}
	return ans;
}
int main() 
{
	int T;cin>>T;
	while(T--)
	{
		int n,m;
		cin>>n>>m;
		rep(i,1,n+1)scanf("%d",a+i);
		build(1,1,n);
		while(m--)
		{
			char op[11];
			int x,y;
			scanf("%s%d%d",op,&x,&y);
			if(op[0] == 'U')
			{
				a[x+1] = y;
				update(x+1,y,1);
			}
			else
			{
				printf("%d\n",query(x+1,y+1,1));
			}
		}
	}
    return 0;
}