线段树(区间合并) HDOJ 3308 LCIS

题目传送门

题意：线段树操作：1. 单点更新 2. 求区间的LCIS(longest consecutive increasing subsequence)

分析：注意是连续的子序列，就是简单的区间合并，记录线段的端点的值，如果rx[rt<<1] < lx[rt<<1|1]就区间合并，最后结果保存在ms中。因为记录的数据较多，索性结构体内再开一个结构体，感觉还不错。写完这题，对区间合并的操作有了更深的理解。

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/25 星期五 10:37:34
* File Name     :G.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
struct ST   {
    struct Node {
        int lx, rx;
        int ls, rs, ms;
    }node[N<<2];
    void push_up(int l, int r, int rt)    {
        node[rt].ls = node[rt<<1].ls;   node[rt].rs = node[rt<<1|1].rs;
        node[rt].lx = node[rt<<1].lx;   node[rt].rx = node[rt<<1|1].rx;
        node[rt].ms = max (node[rt<<1].ms, node[rt<<1|1].ms);
        int mid = (l + r) >> 1;
        if (node[rt<<1].rx < node[rt<<1|1].lx)  {
            if (node[rt<<1].ls == mid - l + 1)  {
                node[rt].ls += node[rt<<1|1].ls;
            }
            if (node[rt<<1|1].rs == r - mid)    {
                node[rt].rs += node[rt<<1].rs;
            }
            node[rt].ms = max (node[rt].ms, node[rt<<1].rs + node[rt<<1|1].ls);
        }
    }
    void build(int l, int r, int rt)    {
        if (l == r) {
            scanf ("%d", &node[rt].lx);
            node[rt].rx = node[rt].lx;
            node[rt].ls = node[rt].rs = node[rt].ms = 1;
            return ;
        }
        int mid = (l + r) >> 1;
        build (lson);   build (rson);
        push_up (l, r, rt);
    }
    void updata(int p, int c, int l, int r, int rt) {
        if (l == r) {
            node[rt].lx = node[rt].rx = c;  return ;
        }
        int mid = (l + r) >> 1;
        if (p <= mid)   updata (p, c, lson);
        else    updata (p, c, rson);
        push_up (l, r, rt);
    }
    int query(int ql, int qr, int l, int r, int rt) {
        if (ql <= l && r <= qr) {
            return node[rt].ms;
        }
        int mid = (l + r) >> 1, ret = 0;
        if (ql <= mid)  {
            ret = max (ret, query (ql, qr, lson));
        }
        if (qr > mid)   {
            ret = max (ret, query (ql, qr, rson));
        }
        if (node[rt<<1].rx < node[rt<<1|1].lx)  {
            ret = max (ret, min (mid - ql + 1, node[rt<<1].rs) + min (qr - mid, node[rt<<1|1].ls));
        }
        return ret;
    }
}st;

int main(void)    {
    int T;  scanf ("%d", &T);
    while (T--)   {
        int n, q;   scanf ("%d%d", &n, &q);
        st.build (1, n, 1);
        for (int a, b, i=1; i<=q; ++i)    {
            char s[2];   scanf ("%s%d%d", &s, &a, &b);
            if (s[0] == 'Q')  {
                printf ("%d
", st.query (a + 1, b + 1, 1, n, 1));
            }
            else    {
                st.updata (a + 1, b, 1, n, 1);
            }
        }
    }

    return 0;
}

新代码

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 1e5 + 5;

#define lc o << 1
#define rc o << 1 | 1

int s[N<<2], sl[N<<2], sr[N<<2];
int al[N<<2], ar[N<<2];
int x[N];

void push_up(int o, int l, int r) {
    sl[o] = sl[lc]; sr[o] = sr[rc];
    al[o] = al[lc]; ar[o] = ar[rc];
    s[o] = max(s[lc], s[rc]);
    int mid = l + r >> 1;
    if (ar[lc] < al[rc]) {
        s[o] = max(s[o], sr[lc]+sl[rc]);
        if (sl[o] == mid-l+1) sl[o] += sl[rc];
        if (sr[o] == r-mid) sr[o] += sr[lc];
    }
}

void build(int o, int l, int r) {
    if (l == r) {
        al[o] = ar[o] = x[l];
        s[o] = sl[o] = sr[o] = 1;
        return ;
    }
    int mid = l + r >> 1;
    build(lc, l, mid);
    build(rc, mid+1, r);
    push_up(o, l, r);
}

void modify(int o, int l, int r, int p, int v) {
    if (l == r) {
        al[o] = ar[o] = v;
        return ;
    }
    int mid = l + r >> 1;
    if (p <= mid) modify(lc, l, mid, p, v);
    else modify(rc, mid+1, r, p, v);
    push_up(o, l, r);
}

int query(int o, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return s[o];
    int mid = l + r >> 1, ret = 0;
    if (ql <= mid) ret = max(ret, query(lc, l, mid, ql, qr));
    if (qr > mid) ret = max(ret, query(rc, mid+1, r, ql, qr));
    if (ql <= mid && qr > mid) {
        if (ar[lc] < al[rc]) {
            ret = max(ret, min(sr[lc], mid-ql+1)+min(sl[rc], qr-mid));
        }
    }
    return ret;
}

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", &x[i]);
        }
        build(1, 1, n);
        char str[5];
        int a, b;
        while (m--) {
            scanf("%s%d%d", &str, &a, &b);
            if (str[0] == 'U') modify(1, 1, n, a+1, b);
            else printf("%d
", query(1, 1, n, a+1, b+1));
        }
    }
    return 0;
}