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

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

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;
}