HDU 3308 LCIS

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本文介绍了解决LCIS（最长连续递增子串）问题的一种高效算法，该算法利用了线段树进行区间合并操作，支持单点更新及查询指定区间内最长连续递增子串的长度。

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LCIS

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5890 Accepted Submission(s): 2555

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

Author

shǎ崽

Source

HDOJ Monthly Contest – 2010.02.06

线段树区间合并

和Hotel那道题有些相似，基本做法是一致的。

题目大意:

一串数字，可以单点更新，可以查询某一区间内的最长单调递增字串（是子串，而不是子序列；字串是连续的，子序列不需要连续）的长度。

注意题目的区间索引是从0开始的。

#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXN=100005;

int num[MAXN];
int left[MAXN<<2],right[MAXN<<2],middle[MAXN<<2];

void pushUp(int x,int l,int r)
{
        left[x]=left[x<<1];
        right[x]=right[x<<1|1];
        middle[x]=max(middle[x<<1],middle[x<<1|1]);
        int center=(l+r)/2;
        int length=r-l+1;
        if(num[center]<num[center+1])
        {
                if(left[x]==length-length/2)
                        left[x]+=left[x<<1|1];
                if(right[x]==length/2)
                        right[x]+=right[x<<1];
                middle[x]=max(right[x<<1]+left[x<<1|1],middle[x]);
        }
}

void build(int x,int l,int r)
{
        if(l==r)
        {
                left[x]=right[x]=middle[x]=1;
                return;
        }
        int center=(l+r)/2;
        build(x<<1,l,center);
        build(x<<1|1,center+1,r);
        pushUp(x,l,r);
}

void update(int x,int var,int l,int r,int p)
{
        if(l==r)
        {
                num[l]=var;
                return;
        }
        int center=(l+r)/2;
        if(p<=center)
                update(x<<1,var,l,center,p);
        else
                update(x<<1|1,var,center+1,r,p);
        pushUp(x,l,r);
}

int query(int x,int l,int r,int L,int R)
{
        if(L<=l && r<=R)
                return middle[x];
        int center=(l+r)/2;
        if(R<=center)
                return query(x<<1,l,center,L,R);
        else if(center<L)
                return query(x<<1|1,center+1,r,L,R);
        else
        {
                int temp = 0;
                if(num[center] < num[center+1])
                        temp = min(right[x<<1],center-L+1) + min(left[x<<1|1],R - center);
                int t1 = query(x<<1,l,center,L,R);
                int t2 = query(x<<1|1,center+1,r,L,R);
                return max(temp,max(t1,t2));
        }
}

int main()
{
        int t,n,m,a,b;
        char option[5];
        scanf("%d",&t);
        while(t--)
        {
                scanf("%d%d",&n,&m);
                for(int i=1;i<=n;i++)
                        scanf("%d",&num[i]);
                build(1,1,n);
                while(m--)
                {
                        scanf("%s",option);
                        scanf("%d%d",&a,&b);
                        if(option[0]=='U')
                                update(1,b,1,n,a+1);
                        else
                                printf("%d\n",query(1,1,n,a+1,b+1));
                }
        }
        return 0;
}