HDOJ 3308 LCIS （线段树之区间合并）

LCIS

[b]Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7349    Accepted Submission(s): 3110
[/b]

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

#include <bits/stdc++.h>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define f(i,a,b) for(int i=(a);i<(b);++i)
const int maxn =10005;
const int mod = 10005;
const int INF = 1e9;
#define ll long long
#define m ((l+r)>>1)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define rush() int T;scanf("%d",&T);while(T--)
int msum[maxn<<2]; //区间内最长连续递增子序列长度.  
int lsum[maxn<<2]; //从第一个数开始算起 到a[m]的最大连续递增序列长度
int rsum[maxn<<2]; //从a[m+1]开始算起 到最后一个数的最大连续递增序列长度
int num[maxn];
void pushup(int l,int r,int rt)
{
lsum[rt]=lsum[rt*2];
rsum[rt]=rsum[rt*2+1];
msum[rt]=max(msum[rt*2],msum[rt*2+1]);
int len=r-l+1;
if(num[m]<num[m+1])
{
if(lsum[rt]==len-(len/2))
lsum[rt]+=lsum[rt*2+1];
if(rsum[rt]==len/2)
rsum[rt]+=rsum[rt*2];
msum[rt]=max(msum[rt],lsum[rt*2+1]+rsum[rt*2]);
}

}
void build(int l,int r,int rt)
{
if(l==r)
{
msum[rt]=lsum[rt]=rsum[rt]=1;
return ;
}
build(lson);
build(rson);
pushup(l,r,rt) ;
}
void update(int p,int l,int r,int rt)
{
if(l==r)
return;
if(p<=m)
update(p,lson);
else
update(p,rson);
pushup(l,r,rt);
}
int query(int x,int y,int l,int r,int rt)
{
if(x<=l&&r<=y)
{
return msum[rt];
}
if(y<=m) return query(x,y,lson);
if(x>m)  return query(x,y,rson);
int a,b;
a=query(x,y,lson);
b=query(x,y,rson);
int ans;
ans=max(a,b);
if(num[m]<num[m+1]) //区间合并
{
int c;
c=min(rsum[rt*2],m-x+1)+min(lsum[rt*2+1],y-m);//防炸
ans=max(c,ans);
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,M;
mst(lsum,0),mst(rsum,0),mst(msum,0),mst(num,0);
scanf("%d%d",&n,&M);
f(i,1,n)
{
scanf("%d",&num[i]);
}
build(1,n,1);
while(M--)
{
char ch[2];
scanf("%s",ch);
int a,b;
scanf("%d%d",&a,&b);
if(ch[0]=='U')
{
a++;
num[a]=b;
update(a,1,n,1);
}
else
{
a++,b++;
printf("%d\n",query(a,b,1,n,1));
}
}
}
return 0;
}