POJ 1754 Segment Tree

E, it should be a water problem in the line tree. Line segment Tree single point update. The most value between Sany.

Code Template PS: Templates Some places don't quite understand.

#include <stdio.h>
#include <iostream>
#include <string.h>
#define MAXN 200010
using namespace Std;

int VAL[MAXN];

struct Node
{
int Max; We need to find the maximum value here.
int left, right;
}TREE[MAXN*4]; Said. The segment tree node needs to be an array of three to five times larger

int max (int a, int b)
{
Return a > B? A:B;
}

int build (int root, int left, int right)//root node left and right array-length tree
{
Tree[root].left = left;
Tree[root].right = right;
if (left = = right)
return Tree[root].max = Val[left]; assigning values to leaf nodes

int mid = (left + right)/2;
int a = Build (2*root, left, mid); Establishment of Saozi right sub-tree
int B = Build (2*root+1, mid+1, right);
return Tree[root].max = Max (A, b);
}

int update (int root, int pos, int val)//single-point update update the POS position element to Val
{
if (Tree[root].left = = pos && tree[root].right = = pos)//If the node corresponding to the location is found
return Tree[root].max = val;
if (Tree[root].left > pos | | tree[root].right < POS)//If not. These are two kinds of recursive breach. E. But I'm not quite sure what the return value of these two cases is.
return Tree[root].max;
int A, B;
A = Update (2*root, POS, Val); Update Saozi Right subtree
b = Update (2*root+1, POS, Val);
return Tree[root].max = Max (A, b);
}

int calculate (int root, int left, int right)//find the maximum value of the interval left to right
{
if (Tree[root].left > right | | tree[root].right < left)//If there is no intersection between the current interval and the range you are seeking
return 0;
if (left <= tree[root].left && tree[root].right <= right)//If the current interval is fully contained within the desired interval. I do not understand why we should return the maximum value of the current interval.
return Tree[root].max;
int A, B;
A = Calculate (2*root, left, right); Both of these cases are wood. We need to decompose the interval to find the maximum value of the sub-interval.
b = Calculate (2*root+1, left, right);
Return Max (A, b);
}

int main ()
{
int n, m;
while (CIN >> n >> m)
{
for (int i=1; i<=n; ++i)
scanf ("%d", &val[i]);
Build (1, 1, N);
while (m--)
{
char temp;
int A, B;
CIN >> Temp >> a >> b;
if (temp = = ' Q ')
cout << Calculate (1, a, b) << Endl;
else update (1, a, b);
}
}
return 0;
}

POJ 1754 Segment Tree