Hdoj1754 I hate it [Line Segment tree]

I hate it Time Limit: 9000/3000 MS (Java/others) memory limit: 32768/32768 K (Java/Others) Total submission (s): 35417 accepted submission (s): 13958

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 row 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 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

This proves that macro definition is slower than custom function. But I don't know why. Then I asked the teacher, and replied, "The function call parameter is calculated only once. The macro definition is not calculated, and the calculation is performed every time during the runtime ."

Run ID	Submit time	Judge status	Pro. ID	EXE. Time	EXE. Memory	Code Len.	Language	Author
10955721	11:20:55	Accepted	1754	1109 Ms	7156 K	1513 B	G ++	Changmu
10955685	11:18:05	Time limit exceeded	1754	3000 Ms	7144 K	1491 B	G ++	Changmu

AC:

#include <stdio.h>#define maxn 200002struct Node{int left, right;int max;} tree[maxn << 2];int stu[maxn];int MAX(int a, int b){return a > b ? a : b;}void build(int left, int right, int rt){tree[rt].left = left; tree[rt].right = right;if(left == right){tree[rt].max = stu[left]; return;}int mid = (left + right) >> 1;build(left, mid, rt << 1);build(mid + 1, right, rt << 1 | 1);tree[rt].max = MAX(tree[rt<<1].max, tree[rt<<1|1].max);}void update(int pos, int val, int rt){if(tree[rt].left == tree[rt].right){tree[rt].max = val; return;}int mid = (tree[rt].left + tree[rt].right) >> 1;if(pos <= mid) update(pos, val, rt << 1);else update(pos, val, rt << 1 | 1);tree[rt].max = MAX(tree[rt<<1].max, tree[rt<<1|1].max);}int query(int left, int right, int rt){if(tree[rt].left == left && right == tree[rt].right){return tree[rt].max;}int mid = (tree[rt].left + tree[rt].right) >> 1;if(right <= mid) return query(left, right, rt << 1);else if(left > mid) return query(left, right, rt << 1 | 1);return MAX(query(left, mid, rt << 1), query(mid+1, right, rt<<1|1));}int main(){int N, M, i, m, n;char com[2];while(scanf("%d%d", &N, &M) == 2){for(i = 1; i <= N; ++i)scanf("%d", stu + i);build(1, N, 1);while(M--){scanf("%s%d%d", com, &m, &n);if(com[0] == 'U') update(m, n, 1);else printf("%d\n", query(m, n, 1));}}return 0;}

TLE ::

#include <stdio.h>#define MAX(a, b) a > b ? a : b#define maxn 200002struct Node{    int lson, rson;    int max;} tree[maxn << 2];int stu[maxn];void build(int left, int right, int rt){    tree[rt].lson = left; tree[rt].rson = right;    if(left == right){        tree[rt].max = stu[left]; return;    }        int mid = (left + right) >> 1;    build(left, mid, rt << 1);    build(mid + 1, right, rt << 1 | 1);        tree[rt].max = MAX(tree[rt<<1].max, tree[rt<<1|1].max);}void update(int pos, int val, int rt){    if(tree[rt].lson == tree[rt].rson){        tree[rt].max = val; return;    }        int mid = (tree[rt].lson + tree[rt].rson) >> 1;    if(pos <= mid) update(pos, val, rt << 1);    else update(pos, val, rt << 1 | 1);        tree[rt].max = MAX(tree[rt<<1].max, tree[rt<<1|1].max);}int query(int left, int right, int rt){    if(tree[rt].lson == left && right == tree[rt].rson){        return tree[rt].max;    }        int mid = (tree[rt].lson + tree[rt].rson) >> 1;    if(right <= mid) return query(left, right, rt << 1);    else if(left > mid)         return query(left, right, rt << 1 | 1);    return MAX(query(left, mid, rt << 1), query(mid+1, right, rt<<1|1));}int main(){    int N, M, i, m, n;    char com[2];    while(scanf("%d%d", &N, &M) == 2){        for(i = 1; i <= N; ++i)            scanf("%d", stu + i);        build(1, N, 1);                while(M--){            scanf("%s%d%d", com, &m, &n);                        if(com[0] == 'U') update(m, n, 1);            else printf("%d\n", query(m, n, 1));        }    }    return 0;}