Description
Many schools are popular for a comparative habit. Teachers like to ask, from XXX to XXX, the highest score is how much.
This makes many students very disgusted.
whether you like it or not, now you need to do is to follow the teacher's request, write a program, mock teacher's inquiry. Of course, teachers sometimes need to update a student's grades.
Input
This topic contains multiple sets of tests, please handle to the end of the file.
On the first line of each test, there are two positive integers N and M (0<n<=200000,0<m<5000), representing the number of students and the number of operations respectively.
Student ID numbers are numbered from 1 to N, respectively.
The second line contains n integers representing the initial scores of the N students, of which the number of I represents the student's score for ID i.
Then there's M-line. Each line has a character C ( only ' Q ' or ' U ') , and two positive integers, A/b.
When C is ' Q ', it indicates that this is a query operation, which asks for the highest number of students whose IDs are from a to B (including A, a).
when C is ' U ', it indicates that this is an update operation that requires the student with ID A to change the grade to B.
Output
for each query operation, output the highest score in one line.
Sample Input
5 6
1 2 3 4 5Q 1 5U 3 6Q 3 4Q 4 5U 2 9Q 1 5
Sample Output
5659
Hint
Cin
Explanation: The main meaning of the problem in how to spend less time to achieve a student's highest score and How to update a student's performance. so this problem, have the following ideas:
1. When constructing each segment tree, in addition to the bottom row, each Fulcrum stores the maximum (not and) of the left child and right child below, so that a student is read
The maximum score will be much faster.
2. Because each time with a new student performance, the above fulcrum must also make corresponding changes, so the use of recursive thinking from the bottom up to update the results.
Here's the code:
1#include <cstdio>2#include <cstring>3#include <iostream>4 #defineMax (A > B) (A:B)5 #defineMin (a > B) (b:a)6 7 using namespacestd;8 9 Const intmaxn=200010;Ten One structStudent A { - intLeft,right; - intNSum; the}segtree[maxn*3]; - intNUM[MAXN]; - - + - + voidBuild (intIintLeftintRight//recursive algorithm to construct a segment tree A { atsegtree[i].left=Left ; -segtree[i].right=Right ; - if(left==Right ) - { -segtree[i].nsum=Num[left]; - return; in } - intMid= (left+right) >>1;//except 2 toBuild (i<<1, Left,mid);//i*2 a +Build (i<<1|1, mid+1, right);//i*2+1 b -Segtree[i].nsum= (segtree[i<<1].nsum>segtree[i<<1|1].nsum?segtree[i<<1].nsum:segtree[i<<1|1].nsum);//Take maximum value the } * $ Panax Notoginseng voidUintIintTleftintb//Update Results - { the + if(segtree[i].left==tleft&&segtree[i].right==tleft) A { thesegtree[i].nsum=b; + return; - } $ intMid= (segtree[i].left+segtree[i].right) >>1; $ - if(tleft<=mid) -U (i<<1, tleft,b); the Else -U (i<<1|1, tleft,b);Wuyi theSegtree[i].nsum= (segtree[i].nsum>b?segtree[i].nsum:b); - Wu } - About $ intQ (intIintLeftintRight ) - { - if(segtree[i].left==left&&segtree[i].right==Right ) - { A returnsegtree[i].nsum; + } the intMiddle= (segtree[i].left+segtree[i].right) >>1; - if(left>middle) $ { the returnQ (i<<1|1, left,right); the } the Else if(right<=middle) the { - returnQ (i<<1, left,right); in } the Else the { About intA=q (i<<1, left,middle); the intB=q (i<<1|1, middle+1, right); the returnA>b?a:b; the } + } - the Bayi intMain () the { the intn,t,x,y; - CharC; - while(SCANF ("%d%d", &n,&t)! =EOF) { the for(intI=1; i<= N; i++) scanf ("%d",&num[i]); theBuild (1,1, n); the while(t--){ theGetChar ();//absorbs the extra line-break keys -scanf"%c%d%d",&c,&x,&y); the the if(C = ='U') theU1, x, y);94 Else theprintf"%d\n", Q (1, x, y)); the } the } 98 return 0; About}
Acm--i Hate It (Evolutionary version of Segment tree)