标签:线段树
//U A B 将第A位置的元素改为B
//Q A B询问[A , B]区间的最长的递增序列
//维护ma_l , ma , ma_r 左边最长,区间最长,右边最长
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std ;
const int maxn = 1e5 + 10 ;
#define left v<<1
#define right v<<1|1
struct node
{
int l , r ;
int ma_l , ma_r , ma ;
}tree[maxn<<2] ;
int h[maxn] ;
int T ;
int n , m ;
void push_up(int v)
{
tree[v].ma_l = tree[left].ma_l ;
tree[v].ma_r = tree[right].ma_r ;
tree[v].ma = max(tree[left].ma , tree[right].ma) ;
if(h[tree[left].r] < h[tree[right].l])
{
tree[v].ma = max(tree[v].ma , tree[left].ma_r + tree[right].ma_l) ;
if(tree[left].r - tree[left].l + 1 == tree[v].ma_l)
tree[v].ma_l += tree[right].ma_l ;
if(tree[right].r - tree[right].l + 1 == tree[v].ma_r)
tree[v].ma_r += tree[left].ma_r ;
}
}
void build(int l , int r , int v)
{
tree[v].l = l ;
tree[v].r = r;
if(l == r)
{
tree[v].ma_l = tree[v].ma = tree[v].ma_r = 1 ;
return ;
}
int mid = (l + r) >> 1 ;
build(l , mid , left) ;
build(mid + 1 , r , right) ;
push_up(v) ;
}
int query(int v , int l , int r)
{
if(l <= tree[v].l && tree[v].r <= r)
return tree[v].ma ;
int mid = (tree[v].l + tree[v].r) >> 1 ;
int ans_1 = 0 , ans_2 = 0 ,ans_3 = 0 ;
if(l <= mid)ans_1 = query(left , l , r) ;
if(r > mid)ans_2 = query(right , l , r) ;
int ans = max(ans_1 , ans_2) ;
if(h[tree[left].r] < h[tree[right].l])
{
ans_3 = min(tree[left].ma_r , tree[left].r - l + 1) + min(tree[right].ma_l , r - tree[right].l + 1) ;
ans = max(ans , ans_3) ;
}
return ans ;
}
void update(int pos , int num ,int v)
{
if(tree[v].l == tree[v].r)
{
h[pos] = num ;
return ;
}
int mid = (tree[v].l + tree[v].r) >> 1 ;
if(pos <= mid)update(pos , num , left);
else if(pos > mid)update(pos ,num ,right) ;
push_up(v) ;
}
int main()
{
// freopen("in.txt" ,"r" ,stdin) ;
scanf("%d" , &T) ;
while(T--)
{
scanf("%d%d" ,&n , &m) ;
for(int i = 1;i <= n;i++)
scanf("%d" , &h[i]) ;
build(1 , n , 1) ;
char ch[10] ;int a , b ;
while(m--)
{
scanf("%s" ,ch) ;
scanf("%d%d" ,&a , &b) ;
if(ch[0] == ‘Q‘)
printf("%d\n" ,query(1 , a + 1, b + 1));
else if(ch[0] == ‘U‘)
update(a+1 , b , 1) ;
}
}
return 0 ;
}
版权声明:本文为博主原创文章,未经博主允许不得转载。
标签:线段树
原文地址:http://blog.csdn.net/cq_pf/article/details/47070567