HUD-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): 8005    Accepted Submission(s): 3414


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
emm题意不说了。。。解释注释里都有。。233333

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 100050;
int n, m;
int num[maxn];
struct SegTree {
int l, r, c;//左右边界以及连续的长度
int ln, rn;//左右边界的值
int ls, rs, ms;//左右最大的连续上升子序列的最大值，整个区间的最大值
}tree[maxn << 2];

void push_up(int i) {
tree[i].ls = tree[i << 1].ls;
tree[i].rs = tree[i << 1 | 1].rs;
tree[i].ln = tree[i << 1].ln;
tree[i].rn = tree[i << 1 | 1].rn;
tree[i].ms = max(tree[i << 1].ms, tree[i << 1 | 1].ms);
if (tree[i << 1].rn < tree[i << 1 | 1].ln) {
//左子树的右边界值小于右子树的左边界值
if (tree[i << 1].ls == tree[i << 1].c)
tree[i].ls += tree[i << 1 | 1].ls;
if (tree[i << 1 | 1].rs == tree[i << 1 | 1].c)
tree[i].rs += tree[i << 1].rs;
tree[i].ms = max(tree[i].ms, tree[i << 1].rs + tree[i << 1 | 1].ls);
}
}

void build(int i, int l, int r) {
tree[i].l = l;
tree[i].r = r;
tree[i].c = r - l + 1;
if (l == r) {
tree[i].ms = tree[i].ls = tree[i].rs = 1;
tree[i].ln = tree[i].rn = num[l];
return;
}
int mid = (l + r) >> 1;
build(i << 1, l, mid);
build(i << 1 | 1, mid + 1, r);
push_up(i);
}

void update(int i, int k, int val) {
if (tree[i].l == k && tree[i].r == k) {
tree[i].ln = tree[i].rn = val;
return;
}
int mid = (tree[i].l + tree[i].r) >> 1;
if (k <= mid)
update(i << 1, k, val);
else
update(i << 1 | 1, k, val);
push_up(i);
}

int query(int i, int b, int e) {
if (tree[i].l >= b && tree[i].r <= e)
return tree[i].ms;
int mid = (tree[i].l + tree[i].r) >> 1, ans = 0;

if (b <= mid)
ans = max(ans, query(i << 1, b, e));
if (e > mid)
ans = max(ans, query(i << 1 | 1, b, e));
if (tree[i << 1].rn < tree[i << 1 | 1].ln)
ans = max(ans, min(mid - b + 1, tree[i << 1].rs) + min(e - mid, tree[i << 1 | 1].ls));
return ans;
}
void print(int i, int l, int r) {
if (l == r) {
cout << tree[i].rn << "  ";
return ;
}
int mid = (l + r) >> 1;
print(i << 1, l, mid);
print(i << 1 | 1, mid + 1, r);
}
int main() {
//freopen("in.txt", "r", stdin);
ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
while (t--) {
cin >> n >> m;
for (int i = 1; i <= n; ++i) {
cin >> num[i];
}
build(1, 1, n);
char ch[5];
int a, b;
for (int i = 0; i < m; ++i) {
cin >> ch >> a >> b;
if (ch[0] == 'Q') {
//题目给的是从0开始的所以都要加1
cout << query(1, a + 1, b + 1) << endl;
} else {
update(1, a + 1, b);
}
}
}
}