Hdu3308 lcis (simple line segment tree)

Lcis

Time Limit: 6000/2000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 2988 accepted submission (s): 1307


Problem descriptiongiven 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].

Inputt 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 ).

Outputfor 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
 

Authorsh

A string of numbers. One operation is to change the number A to B. The other is to query the LICs of the range [a, B.

Question Analysis: This is a type of question. You can maintain the Left and Right consecutive values of the endpoint at the same time. For more information, seeCode:

# Include <iostream> # include <cstdio> # include <cstring> using namespace STD; const int n = 100005; struct node {int LC, RC; // range left and right lcis int LV, RV; // range left and right endpoint value int ans; // range lcis} tree [n <2]; int M, N; int max (int A, int B) {return A> B? A: B;} int min (int A, int B) {return A> B? B: A;} void pushup (INT num, int S, int e) {int ls = num <1; int rs = num <1 | 1; int mid = (S + E)> 1; tree [num]. lv = tree [ls]. LV; tree [num]. rv = tree [RS]. RV; tree [num]. lc = tree [ls]. LC; If (tree [ls]. ans = mid-S + 1 & tree [ls]. RV <tree [RS]. LV) tree [num]. lc + = tree [RS]. LC; tree [num]. rc = tree [RS]. RC; If (tree [RS]. ans = e-Mid & tree [RS]. LV> tree [ls]. RV) tree [num]. RC + = tree [ls]. RC; int CMP = max (tree [ls]. RC, tree [RS]. lc); If (tree [ls]. RV <tree [RS]. LV) CMP = tree [ls]. RC + tree [RS]. LC; tree [num]. ans = max (tree [ls]. ANS, max (tree [RS]. ANS, CMP);} void build (INT num, int S, int e) {If (S = e) {scanf ("% d ", & tree [num]. LV); tree [num]. rv = tree [num]. LV; tree [num]. lc = tree [num]. rc = 1; tree [num]. ans = 1; return;} int mid = (S + E)> 1; build (Num <1, S, mid); Build (Num <1 | 1, mid + 1, e); pushup (Num, S, e);} void Update (INT num, int S, int e, int POs, int Val) {If (S = e) {tree [num]. lv = tree [num]. rv = val; return;} int mid = (S + E)> 1; if (Pos <= mid) Update (Num <1, S, mid, POs, val); else Update (Num <1 | 1, Mid + 1, E, POs, Val); pushup (Num, S, e);} int query (INT num, int S, int e, int L, int R) {If (S = L & R = e) {return tree [num]. ans;} int mid = (S + E)> 1; if (r <= mid) return query (Num <1, S, mid, L, R ); else {If (L> mid) return query (Num <1 | 1, Mid + 1, E, L, R ); else {int TA = min (tree [num <1]. RC, mid-L + 1); int TB = min (tree [num <1 | 1]. LC, R-mid); int middle = max (TA, Tb); If (tree [num <1]. RV <tree [num <1 | 1]. LV) Middle = Ta + Tb; return max (query (Num <1, S, mid, L, mid), max (query (Num <1 | 1, mid + 1, E, Mid + 1, R), middle) ;}} int main () {int t; char op [3]; int A, B; scanf ("% d", & T); While (t --) {scanf ("% d", & N, & M); Build (1, 1, n ); while (M --) {scanf ("% s", OP); scanf ("% d", & A, & B ); if (OP [0] = 'q') printf ("% d \ n", query (1, 1, n, A + 1, B + 1 )); else Update (, N, A + 1, B) ;}} return 0 ;}// 343ms5376k