HDU 3308 线段树单点更新+区间查找最长连续子序列

LCIS

Time Limit: 6000/2000 MS (Java/Others)

Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5591 Accepted Submission(s): 2443


[align=left]Problem Description[/align]
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].

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each Q, output the answer.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

1
1
4
2
3
1
2
5

[align=left]Author[/align]
shǎ崽

题解：开始看错题，以为是最长子序列，GG，
后知后觉，线段树秒了

///1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a))
#define memfy(a)  memset(a,-1,sizeof(a))
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b) scanf("%d%d",&a,&b)
#define mod 1000000007
#define inf 1000000001
#define maxn 200000+2
inline ll read()
{
ll x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
//****************************************
struct ss
{
int l,r,v,ans;
int lc,rc;
}tr[maxn<<2];
int a[maxn<<2],n,q;
void push_up(int k)
{
if(a[tr[k<<1].r]<a[tr[k<<1|1].l])
{
tr[k].ans=max(max(tr[k<<1].ans,tr[k<<1|1].ans),tr[k<<1].rc+tr[k<<1|1].lc);
if(tr[k<<1].lc==(tr[k<<1].r-tr[k<<1].l+1))
{
tr[k].lc=tr[k<<1].lc+tr[k<<1|1].lc;
}else  tr[k].lc=tr[k<<1].lc;
if(tr[k<<1|1].lc==(tr[k<<1|1].r-tr[k<<1|1].l+1))
{
tr[k].rc=tr[k<<1|1].rc+tr[k<<1].rc;
}else  tr[k].rc=tr[k<<1|1].rc;
}
else
{
tr[k].ans=max(tr[k<<1].ans,tr[k<<1|1].ans);
tr[k].lc=tr[k<<1].lc;
tr[k].rc=tr[k<<1|1].rc;
}
}
void build(int k,int s,int t)
{
tr[k].l=s;
tr[k].r=t;
if(s==t)
{
tr[k].v=a[s];
tr[k].ans=1;
tr[k].lc=1;
tr[k].rc=1;
return ;
}
int mid=(s+t)>>1;
build(k<<1,s,mid);
build(k<<1|1,mid+1,t);
push_up(k);
}
void update(int k,int x,int c)
{
if(tr[k].l==x&&tr[k].r==x)
{
tr[k].v=c;
a[x]=c;
return ;
}
int mid=(tr[k].l+tr[k].r)>>1;
if(x<=mid)update(k<<1,x,c);
else update(k<<1|1,x,c);
push_up(k);
}
int ask(int k,int s,int t)
{
if(tr[k].l==s&&tr[k].r==t)
{
return tr[k].ans;
}
int mid=(tr[k].l+tr[k].r)>>1;
if(t<=mid)return ask(k<<1,s,t);
else if(s>mid)return ask(k<<1|1,s,t);
else {
int ret=0;
int A=ask(k<<1,s,mid);
int B=ask(k<<1|1,mid+1,t);
ret=max(A,B);
A=min(tr[k<<1].rc,mid-s+1);
B=min(tr[k<<1|1].lc,t-mid);
if(a[tr[k<<1].r]<a[tr[k<<1|1].l])
ret=max(A+B,ret);
return ret;
}
}
int main()
{

int T=read();
while(T--)
{
n=read();
q=read();
FOR(i,1,n)
{
a[i]=read();
}
build(1,1,n);
int a,b;
char ch[321];
FOR(i,1,q)
{
scanf("%s%d%d",ch,&a,&b);
if(ch[0]=='Q')
{
printf("%d\n",ask(1,a+1,b+1));
}
else update(1,a+1,b);
}
}
return 0;
}

代码