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线段树的区间合并 B - LCIS

时间:2019-05-03 11:27:10      阅读:185      评论:0      收藏:0      [点我收藏+]

标签:+=   asi   stream   dex   div   else   rom   line   while   

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]. 

InputT 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). 
OutputFor 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

 

 

 

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1e5 + 100;
struct node
{
    int l, r, rp, lp, len;
    int max_pre, max_sub, max_last;
}tree[maxn*4];
int a[maxn];

void push_up(int id)
{
    tree[id].rp = tree[id<<1|1].rp;
    tree[id].lp = tree[id << 1].lp;
    tree[id].max_pre = tree[id << 1].max_pre;
    tree[id].max_last = tree[id << 1 | 1].max_last;
    tree[id].max_sub = max(tree[id << 1].max_sub, tree[id << 1 | 1].max_sub);
    if(tree[id<<1].rp<tree[id<<1|1].lp)
    {
        if (tree[id << 1].max_pre == tree[id << 1].len) tree[id].max_pre += tree[id << 1 | 1].max_pre;
        if (tree[id << 1 | 1].max_last == tree[id << 1 | 1].len) tree[id].max_last += tree[id << 1].max_last;
        tree[id].max_sub = max(tree[id].max_sub, tree[id << 1].max_last + tree[id<<1|1].max_pre);
    }
}

void build(int id,int l,int r)
{
    tree[id].l = l;
    tree[id].r = r;
    tree[id].len = r - l + 1;
    if(l==r)
    {
        tree[id].rp = tree[id].lp = a[l];
        tree[id].max_pre = tree[id].max_last = tree[id].max_sub = 1;
        return;
    }
    int mid = (l+r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
    push_up(id);
}

int query(int id,int x,int y)
{
    int ans = 0;
    if(x<=tree[id].l&&tree[id].r<=y)
    {
        return tree[id].max_sub;
    }
    int mid = (tree[id].l + tree[id].r) >> 1;
    if (x <= mid) ans = max(ans, query(id << 1, x, y));
    if (y > mid) ans = max(ans, query(id << 1 | 1, x, y));
    if(tree[id<<1].rp<tree[id<<1|1].lp)
    {
        ans = max(ans, min(mid - x + 1, tree[id << 1].max_last) + min(y - mid, tree[id << 1 | 1].max_pre));
    }
    return ans;
}

void update(int id,int x,int z)
{
    if(tree[id].l==tree[id].r&&tree[id].l==x)
    {
        tree[id].rp = z;
        tree[id].lp = z;
        return;
    }
    int mid = (tree[id].l + tree[id].r) >> 1;
    if (x <= mid) update(id << 1, x, z);
    if (x > mid) update(id << 1 | 1, x, z);
    push_up(id);
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, m;
        scanf("%d%d", &n, &m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d", &a[i]);
        }
        build(1, 1, n);
        while(m--)
        {
            char s[10];
            int x, y;
            scanf("%s %d %d", s, &x, &y);
            if(s[0]==Q)
            {
                int ans = query(1, x+1, y+1);
                printf("%d\n", ans);
            }
            else update(1, x+1, y);
        }
    }
}

 

线段树的区间合并 B - LCIS

标签:+=   asi   stream   dex   div   else   rom   line   while   

原文地址:https://www.cnblogs.com/EchoZQN/p/10804394.html

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