hdu-3308 LCIS (线段树区间合并）

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8337    Accepted Submission(s): 3566

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<=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

 

思路：

n个数字，q个操作，有两种操作：1.Q询问操作：在x,y区间内最长的连续递增子序列的长度。2，替换操作，把下标为x数替换为y.

这种询问，单点更新操作一般都是用线段树做。之前一直想成了最长递增子序列。。。完全没有思路，，其实他只要求最长的连续子串长度。。注意：是最长的连续的子串。

这样就是很裸的线段树区间合并了，

 

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 2e5+20;
int lsum[M<<2],sum[M<<2],rsum[M<<2],num[M];
void pushup(int l,int r,int rt){
    int k = r - l + 1;
    mid;
    lsum[rt] = lsum[rt<<1]; rsum[rt] = rsum[rt<<1|1];
    sum[rt] = max(sum[rt<<1],sum[rt<<1|1]);
    if(num[m] < num[m+1]){
        if(lsum[rt] == (k - (k >> 1))) lsum[rt] += lsum[rt<<1|1];
        if(rsum[rt] == (k>>1)) rsum[rt] += rsum[rt<<1];
        sum[rt] = max(sum[rt],lsum[rt<<1|1]+rsum[rt<<1]);
    }
}
void build(int l,int r,int rt){
     if(l == r){
        sum[rt] = lsum[rt] = rsum[rt]  = 1;
        return ;
     }
     mid;
     build(lson);
     build(rson);
     pushup(l,r,rt);
}
void update(int p,int c,int l,int r,int rt){
    if(l == r){
        num[l] = c;
        return ;
    }
    mid;
    if(p <= m) update(p,c,lson);
    if(p > m) update(p,c,rson);
    pushup(l,r,rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L <= l&&R >= r) return sum[rt];
    int ret = 0;
    mid;
    if(L <= m) ret = max(ret,query(L,R,lson));
    if(R > m) ret = max(ret,query(L,R,rson));
    if(num[m] < num[m+1]){
        ret = max(ret,min(m - L + 1,rsum[rt<<1])+min(R-m,lsum[rt<<1|1]));
    }
    return ret;
}

int main(){
    int t,n,q,x,y;
    char c;
    scanf("%d",&t);
    while(t--){
        cin>>n>>q;
        for(int i = 1; i <= n;i ++){
            cin>>num[i];
        }
        build(1,n,1);
        for(int i = 0;i < q;i ++){
            cin>>c>>x>>y;
            if(c == 'Q'){
                x++;y++;
                cout<<query(x,y,1,n,1)<<endl;
            }
            else{
                x++;
                update(x,y,1,n,1);
            }
        }
    }
    return 0;
}