Hdu1754 I hate it (line segment tree)

Hdu1754 I hate it (line segment tree)
I Hate ItTime Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission (s): 55291 Accepted Submission (s): 21599

Problem Description many schools have a popular habit. Teachers like to ask, from xx to xx, what is the highest score.
This made many students very disgusted.

Whether you like it or not, what you need to do now is to write a program to simulate the instructor's inquiry according to the instructor's requirements. Of course, teachers sometimes need to update their scores.
Input this question contains multiple groups of tests, please process until the end of the file.
In the first line of each test, there are two positive integers N and M (0 Student ID numbers are separated from 1 to N.
The second row contains N integers, indicating the initial score of the N students. The number of I represents the score of the students whose ID is I.
Next there are M rows. Each line has A character C (only 'q' or 'U'), and two positive integers A and B.
When C is 'Q', it indicates that this is A query operation. It asks the students whose ID ranges from A to B (including A and B) about the highest score.
When C is 'U', it indicates that this is an update operation. You must change the score of students whose ID is A to B.

Output outputs the highest score in one row for each query operation.
Sample Input

5 61 2 3 4 5Q 1 5U 3 6Q 3 4Q 4 5U 2 9Q 1 5

Sample Output

5659HintHuge input,the C function scanf() will work better than cin 

Author linle
Analysis: basic question for getting started with the line segment tree. Each time a node is updated, the maximum value of the node-based subtree is stored.

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          #include using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 0x3f3f3f3f;const int MOD = 1000000007;#define ll long long#define CL(a,b) memset(a,b,sizeof(a))#define MAXN 800010struct node{ int a,b,r;}t[MAXN];int n,m,ans;void build(int x, int y, int num){ t[num].a = x; t[num].b = y; t[num].r = 0; if(x == y) return ; int mid = (x+y)/2; build(x, mid, num*2); build(mid+1, y, num*2+1);}void update(int x, int y, int num){ if (t[num].a == t[num].b && t[num].b == x) { t[num].r = y; return ; } int mid = (t[num].a+t[num].b)/2; if(x > mid) update(x, y, num*2+1); else update(x, y, num*2); t[num].r = max(t[num*2].r, t[num*2+1].r);}void query(int x, int y, int num){ if(t[num].a == x && t[num].b == y) { ans = max(ans, t[num].r); return ; } int mid = (t[num].a+t[num].b)/2; if(x >= mid+1) query(x, y, num*2+1); else if(y <= mid) query(x, y, num*2); else { query(x, mid, num*2); query(mid+1, y, num*2+1); }}int main(){ char ch; int x,y,k; while(scanf("%d%d",&n,&m)==2) { build(1, n, 1); for(int i=1; i<=n; i++) { scanf("%d",&k); update(i, k, 1); } while(m--) { getchar(); scanf("%c%d%d",&ch,&x,&y); if(ch == 'U') { update(x, y, 1); } else if(ch == 'Q') { ans = -99999; query(x, y, 1); printf("%d\n",ans); } } } return 0;}