线段树---HDU1754 I hate it

这个题也是线段树的基础题，有了上一个题的基础，在做这个题就显得比较轻松了，大体都是一样的，那个是求和，这个改成求最大值，基本上思路差不多，下面是代码的实现

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 200010 * 4;
int segment[MAX];
//向上调整
void pushUp(int root)
{
    segment[root] = max(segment[root * 2], segment[root * 2 + 1]);
}

void buildTree(int root, int left, int right)
{
    if(left == right)
    {
        scanf("%d", &segment[root]);
        return;
    }
    int mid = (left + right) / 2;
    buildTree(root * 2, left, mid);
    buildTree(root * 2 + 1, mid + 1, right);
    //要把跟他上面所有关联的节点都要更新
    pushUp(root);
}
//更新节点
void update(int root, int pos, int update_num, int left, int right)
{
    if(left == right)
    {
        segment[root] = update_num;
        return;
    }
    int mid = (left + right) / 2;
    if(pos <= mid)
    {
        update(root * 2, pos, update_num, left, mid);
    }
    else
    {
        update(root * 2 + 1, pos, update_num, mid + 1, right);
    }
    pushUp(root);
}
//left和right为查到的区间，L和R为需要查询的区间
int getMax(int root, int left, int right, int L, int R)
{
    if(L == left && R == right)
    {
        return segment[root];
    }
    int mid = (left + right) / 2;
    int Max_Num = 0;
    if(R <= mid)
    {
        Max_Num = getMax(root * 2, left, mid, L, R);
    }
    else if(L > mid)
    {
        Max_Num = getMax(root * 2 + 1, mid + 1, right, L, R);
    }
    else
    {
        Max_Num = getMax(root * 2, left, mid, L, mid);
        Max_Num = max(Max_Num, getMax(root * 2 + 1, mid + 1, right, mid + 1, R));
    }
    return Max_Num;
}

int main()
{
    int N, M;
    while(~scanf("%d %d", &N, &M))
    {
        memset(segment, 0, sizeof(segment));
        buildTree(1, 1, N);
        char ch;
        int t1, t2;
        for(int i = 0; i < M; i++)
        {
            getchar();
            scanf("%c %d %d", &ch, &t1, &t2);
            if(ch == 'U')
            {
                update(1, t1, t2, 1, N);
            }
            else
            {
                printf("%d
", getMax(1, 1, N, t1, t2));
            }
        }
    }
    return 0;
}