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].
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).
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.
题目大意：给n个数，动态地修改某个数的值，或者查询一段区间的LCIS（最长连续上升子序列，坑爹的标题……）。
思路：线段树，每个点维护一个区间的从左边开始的最长的LCIS，从右边开始的最长的LCIS，这个区间最大的LCIS。然后随便搞搞就能过了……
代码（593MS）：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 7 const int MAXN = 100010 << 2; 8 9 int lmax[MAXN], rmax[MAXN], mmax[MAXN]; 10 int a[MAXN], n, m, T; 11 12 void update(int x, int l, int r, int pos, int val) { 13 if(pos <= l && r <= pos) { 14 a[pos] = val; 15 } else { 16 int ll = x << 1, rr = ll | 1, mid = (l + r) >> 1; 17 if(pos <= mid) update(ll, l, mid, pos, val); 18 if(mid < pos) update(rr, mid + 1, r, pos, val); 19 if(a[mid] < a[mid + 1]) { 20 lmax[x] = lmax[ll] + (lmax[ll] == mid - l + 1) * lmax[rr]; 21 rmax[x] = rmax[rr] + (rmax[rr] == r - mid) * rmax[ll]; 22 mmax[x] = max(rmax[ll] + lmax[rr], max(mmax[ll], mmax[rr])); 23 } else { 24 lmax[x] = lmax[ll]; 25 rmax[x] = rmax[rr]; 26 mmax[x] = max(mmax[ll], mmax[rr]); 27 } 28 } 29 } 30 31 int query(int x, int l, int r, int aa, int bb) { 32 if(aa <= l && r <= bb) { 33 return mmax[x]; 34 } else { 35 int ll = x << 1, rr = ll | 1, mid = (l + r) >> 1; 36 int ans = 0; 37 if(aa <= mid) ans = max(ans, query(ll, l, mid, aa, bb)); 38 if(mid < bb) ans = max(ans, query(rr, mid + 1, r, aa, bb)); 39 if(a[mid] < a[mid + 1]) ans = max(ans, min(rmax[ll], mid - aa + 1) + min(lmax[rr], bb - mid)); 40 return ans; 41 } 42 } 43 44 void build(int x, int l, int r) { 45 if(l == r) { 46 lmax[x] = rmax[x] = mmax[x] = 1; 47 } else { 48 int ll = x << 1, rr = ll | 1, mid = (l + r) >> 1; 49 build(ll, l, mid); 50 build(rr, mid + 1, r); 51 if(a[mid] < a[mid + 1]) { 52 lmax[x] = lmax[ll] + (lmax[ll] == mid - l + 1) * lmax[rr]; 53 rmax[x] = rmax[rr] + (rmax[rr] == r - mid) * rmax[ll]; 54 mmax[x] = max(rmax[ll] + lmax[rr], max(mmax[ll], mmax[rr])); 55 } else { 56 lmax[x] = lmax[ll]; 57 rmax[x] = rmax[rr]; 58 mmax[x] = max(mmax[ll], mmax[rr]); 59 } 60 } 61 } 62 63 int main() { 64 scanf("%d", &T); 65 while(T--) { 66 scanf("%d%d", &n, &m); 67 for(int i = 0; i < n; ++i) scanf("%d", &a[i]); 68 build(1, 0, n - 1); 69 while(m--) { 70 char c; 71 int a, b; 72 scanf(" %c%d%d", &c, &a, &b); 73 if(c == 'Q') printf("%d ", query(1, 0, n - 1, a, b)); 74 else update(1, 0, n - 1, a, b); 75 } 76 } 77 }