hdu.3308 LCIS(线段树,区间合并+单点更新)

按照傻崽大神的线段树修炼路线，自己做的第二道区间合并的题。

问题比较简单明了，区间求最长连续上升子序列，但是是需要单点更新的

n个数， m组操作

Q A B 询问[A,B]区间的最长连续上升子序列；

O A B 把位置为A的数字改成B。(位置从0开始)

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4291    Accepted Submission(s): 1958

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

110 107 7 3 3 5 9 9 8 1 8 Q 6 6U 3 4Q 0 1Q 0 5Q 4 7Q 3 5Q 0 2Q 4 6U 6 10Q 0 9

 

Sample Output

11423125

 

Author

shǎ崽

 

线段树上的每个节点记录的信息有：

1.左右两端的端点值(用于鉴定左右两个区间能否合并)

2.左、右儿子的最长连续上升子序列长度，总区间的最长连续上升子序列长度。

PS：查询的时候需要判断下区间长度和连续的长度哪个小，以免连续长度超过区间长度。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define LL(x) (x << 1)#define RR(x) (x << 1 | 1)#define MID(x, y) (x + ((y - x) >> 1))#define lson l, mid, LL(t)#define rson mid + 1, r, RR(t)const int MAXN = 100100;int t, n, m;int a[MAXN];struct node {    int lmx, rmx, mx;//左、右和总体最长公共上升子序列长度    int lp, rp;//左右端点值};class segment_tree {    public:    node tree[MAXN << 2];    int point[MAXN << 2];    void pushup(int t) {        tree[t].lp = tree[LL(t)].lp; tree[t].rp = tree[RR(t)].rp;        tree[t].lmx = tree[LL(t)].lmx; tree[t].rmx = tree[RR(t)].rmx;        tree[t].mx = max(tree[LL(t)].mx, tree[RR(t)].mx);        if(tree[LL(t)].rp < tree[RR(t)].lp) {            tree[t].mx = max(tree[t].mx, tree[LL(t)].rmx + tree[RR(t)].lmx);            if(tree[LL(t)].lmx == point[LL(t)])                tree[t].lmx = tree[LL(t)].lmx + tree[RR(t)].lmx;            if(tree[RR(t)].rmx == point[RR(t)])                tree[t].rmx = tree[RR(t)].rmx + tree[LL(t)].rmx;        }        tree[t].mx = max(tree[t].mx, max(tree[t].lmx, tree[t].rmx));    }    void build(int l, int r, int t) {        if(l == r) {            tree[t].lp = tree[t].rp = a[l];            tree[t].lmx = tree[t].rmx = tree[t].mx = 1;            point[t] = 1;        } else {            int mid = MID(l, r);            build(lson);            build(rson);            point[t] = point[LL(t)] + point[RR(t)];            pushup(t);        }    }    int query(int st, int ed, int l, int r, int t) {        if(st <= l && r <= ed) return tree[t].mx;        else {            int mid = MID(l, r);            if(ed <= mid) return query(st, ed, lson);            else if(st > mid) return query(st, ed, rson);            else {                int temp1 = 0, temp2 = 0;                int mx1 = query(st, ed, lson);                int mx2 = query(st, ed, rson);                if(tree[LL(t)].rp < tree[RR(t)].lp) {                    temp1 = min(mid - st + 1, tree[LL(t)].rmx);                    temp2 = min(ed - mid, tree[RR(t)].lmx);                }                return max(max(mx1 , mx2), temp1 + temp2);            }        }    }    void update(int pos, int l, int r, int t, int w) {        if(l == r) tree[t].lp = tree[t].rp = w;        else {            int mid = MID(l, r);            if(pos <= mid) update(pos, lson, w);            else update(pos, rson, w);            pushup(t);        }    }}seg;int main() {    scanf("%d", &t);    while(t--) {        scanf("%d%d", &n, &m);        memset(a, 0, sizeof(a));        for(int i = 0; i < n; i++)            scanf("%d", &a[i]);        seg.build(0, n - 1, 1);        char ch[2]; int x, y;        while(m--) {            scanf("%s%d%d", ch, &x, &y);            if(ch[0] == 'Q') {                printf("%d\n", seg.query(x, y, 0, n - 1, 1));            } else {                seg.update(x, 0, n - 1, 1, y);            }        }    }    return 0;}