LCIS
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3717 Accepted Submission(s): 1660
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
HDOJ Monthly Contest – 2010.02.06
传送门：HDU 3308 LCIS
题目大意：有两种操作，1、修改单点的值，2、询问一段区间内的连续上升序列LCIS。
题目分析：就是维护左端点，右端点，左右连续最大LCIS以及区间内最大LCIS，其他没了。
传送门：HDU 3308 LCIS
题目大意：有两种操作，1、修改单点的值，2、询问一段区间内的连续上升序列LCIS。
题目分析：就是维护左端点，右端点，左右连续最大LCIS以及区间内最大LCIS，其他没了。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define ll o << 1
#define rr o << 1 | 1
#define lson l , m , ll
#define rson m + 1 , r , rr
const int maxN = 1000000 ;
int lnum[maxN] , rnum[maxN] ;
int lmax[maxN] , rmax[maxN] , mmax[maxN] ;
int max ( const int X , const int Y ) {
if ( X > Y ) return X ;
return Y ;
}
int min ( const int X , const int Y ) {
if ( X < Y ) return X ;
return Y ;
}
void PushUp ( int o , int len ) {
lnum[o] = lnum[ll] ;
rnum[o] = rnum[rr] ;
lmax[o] = lmax[ll] ;
rmax[o] = rmax[rr] ;
mmax[o] = max ( mmax[ll] , mmax[rr] ) ;
if ( rnum[ll] < lnum[rr] ) {
mmax[o] = max ( mmax[o] , rmax[ll] + lmax[rr] ) ;
if ( lmax[ll] == len - ( len >> 1 ) ) {
lmax[o] += lmax[rr] ;
}
if ( rmax[rr] == ( len >> 1 ) ) {
rmax[o] += rmax[ll] ;
}
}
}
void Build ( int l , int r , int o ) {
if ( l == r ) {
int x ;
scanf ( "%d" , &x ) ;
lnum[o] = rnum[o] = x ;
lmax[o] = mmax[o] = rmax[o] = 1 ;
return ;
}
int m = ( l + r ) >> 1 ;
Build ( lson ) ;
Build ( rson ) ;
PushUp ( o , r - l + 1 ) ;
}
void Update ( int x , int v , int l , int r , int o ) {
if ( l == r ) {
lnum[o] = rnum[o] = v ;
lmax[o] = mmax[o] = rmax[o] = 1 ;
return ;
}
int m = ( l + r ) >> 1 ;
if ( x <= m ) Update ( x , v , lson ) ;
else Update ( x , v , rson ) ;
PushUp ( o , r - l + 1 ) ;
}
int Query ( int L , int R , int l , int r , int o ) {
if ( L <= l && r <= R ) return mmax[o] ;
int m = ( l + r ) >> 1 ;
int tmp1 = 0 , tmp2 = 0 , tmp3 = 0 ;
if ( L <= m ) tmp1 = Query ( L , R , lson ) ;
if ( m < R ) tmp2 = Query ( L , R , rson ) ;
if ( L <= m && m < R && rnum[ll] < lnum[rr] ) tmp3 = min ( m - L + 1 , rmax[ll] ) + min ( lmax[rr] , R - m ) ;
return max ( max ( tmp1 , tmp2 ) , tmp3 ) ;
}
void work () {
int n , m , First , Second ;
char ch[5] ;
scanf ( "%d%d" , &n , &m ) ;
Build ( 1 , n , 1 ) ;
while ( m -- ) {
scanf ( "%s%d%d" , ch , &First , &Second ) ;
if ( ch[0] == 'U' ) Update ( First + 1 , Second , 1 , n , 1 ) ;
else printf ( "%d\n" , Query ( First + 1 , Second + 1 , 1 , n , 1 ) ) ;
}
}
int main () {
int T ;
scanf ( "%d" , &T ) ;
while ( T -- ) {
work () ;
}
return 0 ;
}