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

 void buildTree(int root, int left, int right)
 {
     if(left == right)
     {
         scanf("%d", &segment[root]);
         return;
     }
     int mid = (left + right) / ;
     buildTree(root * , left, mid);
     buildTree(root *  + , mid + , 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) / ;
     if(pos <= mid)
     {
         update(root * , pos, update_num, left, mid);
     }
     else
     {
         update(root *  + , pos, update_num, mid + , 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) / ;
     int Max_Num = ;
     if(R <= mid)
     {
         Max_Num = getMax(root * , left, mid, L, R);
     }
     else if(L > mid)
     {
         Max_Num = getMax(root *  + , mid + , right, L, R);
     }
     else
     {
         Max_Num = getMax(root * , left, mid, L, mid);
         Max_Num = max(Max_Num, getMax(root *  + , mid + , right, mid + , R));
     }
     return Max_Num;
 }

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