hdu3308 LCIS(简单线段树)

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2988    Accepted Submission(s): 1307

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<=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.

 

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

题目大意：给一串数字，1种操作，将第a个数改成b；一种查询，查询区间[a,b]的LICS。

题目分析：跟这种题一个类型，同时维护端点左右连续值就可以了。详情请见代码：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 100005;
struct node
{
    int lc,rc;//区间左右LCIS
    int lv,rv;//区间左右端点值
    int ans;//区间LCIS
}tree[N<<2];
int m,n;

int Max(int a,int b)
{
    return a > b?a:b;
}

int Min(int a,int b)
{
    return a > b?b:a;
}

void pushup(int num,int s,int e)
{
    int ls = num<<1;
    int rs = num<<1|1;
    int mid = (s + e)>>1;
    tree[num].lv = tree[ls].lv;
    tree[num].rv = tree[rs].rv;
    tree[num].lc = tree[ls].lc;
    if(tree[ls].ans == mid - s + 1 && tree[ls].rv < tree[rs].lv)
        tree[num].lc += tree[rs].lc;
    tree[num].rc = tree[rs].rc;
    if(tree[rs].ans == e - mid && tree[rs].lv > tree[ls].rv)
        tree[num].rc += tree[ls].rc;
    int cmp = Max(tree[ls].rc,tree[rs].lc);
    if(tree[ls].rv < tree[rs].lv)
        cmp = tree[ls].rc + tree[rs].lc;
    tree[num].ans = Max(tree[ls].ans,Max(tree[rs].ans,cmp));
}

void build(int num,int s,int e)
{
    if(s == e)
    {
        scanf("%d",&tree[num].lv);
        tree[num].rv = tree[num].lv;
        tree[num].lc = tree[num].rc = 1;
        tree[num].ans = 1;
        return;
    }
    int mid = (s + e)>>1;
    build(num<<1,s,mid);
    build(num<<1|1,mid + 1,e);
    pushup(num,s,e);
}

void update(int num,int s,int e,int pos,int val)
{
    if(s == e)
    {
        tree[num].lv = tree[num].rv = val;
        return;
    }
    int mid = (s + e)>>1;
    if(pos <= mid)
        update(num<<1,s,mid,pos,val);
    else
        update(num<<1|1,mid + 1,e,pos,val);
    pushup(num,s,e);
}

int query(int num,int s,int e,int l,int r)
{
    if(s == l && r == e)
    {
        return tree[num].ans;
    }
    int mid = (s + e)>>1;
    if(r <= mid)
        return query(num<<1,s,mid,l,r);
    else
    {
        if(l > mid)
            return query(num<<1|1,mid + 1,e,l,r);
        else
        {
            int ta = Min(tree[num<<1].rc,mid - l + 1);
            int tb = Min(tree[num<<1|1].lc,r - mid);
            int middle = Max(ta,tb);
            if(tree[num<<1].rv < tree[num<<1|1].lv)
                middle = ta + tb;
            return Max(query(num<<1,s,mid,l,mid),Max(query(num<<1|1,mid + 1,e,mid + 1,r),middle));
        }
    }
}



int main()
{
    int t;
    char op[3];
    int a,b;
    scanf("%d",&t);
    while(t --)
    {
        scanf("%d%d",&n,&m);
        build(1,1,n);
        while(m --)
        {
            scanf("%s",op);
            scanf("%d%d",&a,&b);
            if(op[0] == 'Q')
                printf("%d\n",query(1,1,n,a + 1,b + 1));
            else
                update(1,1,n,a + 1,b);
        }
    }
    return 0;
}
//343MS	5376K