[Line segment tree] [HDU1754] I Hate It (interval maximum value, single-point update)

Link: http://acm.hdu.edu.cn/showproblem.php?pid=1754

 

Single point update, find the maximum value of the interval

  1 // https://blog.csdn.net/panyanyany/article/details/6776300
  2 #define _CRT_SBCURE_NO_DEPRECATE
  3 #include <cmath>
  4 #include <string>
  5 #include <cstdio>
  6 #include <cstdlib>
  7 #include <iostream>
  8 using namespace std;
  9 
 10 const int N = 200010;
 11 
 12 int num[N];
 13 struct node
 14 {
 15     int l,r;
 16     int mmax;    // This question only asks the operation of the highest grade, so no storage interval and 
17 are required      int mid()
 18      {
 19          return (l+r)/ 2 ;
 20      }
 21 }tree[N* 4 ];
 22  
23  // Build a segment tree 
24  int BuildTree( int root, int l, int r)
 25  {
 26      // The interval represented by the current node 
27      tree[root].l = l;
 28      tree[root].r = r;
 29      //If the left and right intervals are the same, this node is a leaf node, and mmax wins to store the value corresponding to a certain student 
30      if (l == r)
 31      {
 32          return tree[root].mmax = num[l];
 33      }
 34      // Recursively build left and right subtrees and get max value from subtree 
35      int a,b;
 36      a = BuildTree(root* 2 ,l,(l+r)/ 2 );
 37      b = BuildTree(root* 2 + 1 ,(l+r)/ 2 + 1 ,r);
 38  
39      return tree[root].mmax = max(a,b);
 40  }
 41  
42 // void add(int root,int i,int t)
 43  // {
 44  //      // tree[root].sum += t;
 45  //      if(tree[root].l == i && tree[ root].r == i)
 46  //          return ;
 47  //      if(i <= tree[root].mid())
 48  //          add(root*2,i,t);
 49  //      else
 50  / /          add(root*2+1,i,t);
 51  // }
 52  
53  
54  
55  
56  // Start from node root, find the maximum value between l and r 
57  int find( introot, int l, int r)
 58  {
 59      // If this interval has no intersection with the interval managed by root 
60      if (tree[root].l > r || tree[root].r < l)
 61          return  0 ;
 62      // If this interval includes the interval managed by root 
63      if (l <= tree[root].l && tree[root].r <= r)
 64          return tree[root].mmax;
 65      // Recursive search left and right Subtree, take the maximum value of 
66      int a,b;
 67      a = find(root* 2 ,l,r);
 68      b = find(root* 2 + 1,l,r);
 69      return max(a,b);
 70  }
 71  // Update the value of pos point 
72  int update( int root, int pos, int val)
 73  {
 74      // If pos does not exist at root Within the managed interval 
75      if (pos < tree[root].l || tree[root].r < pos)
 76          return tree[root].mmax;
 77  
78      // If root happens to be a qualified leaf 
79      if (tree[root].l == pos && tree[root].r == pos)
 80          return tree[root].mmax = val;
81      
82      // else... 
83      int a,b;
 84      a = update(root* 2 ,pos,val);
 85      b = update(root* 2 + 1 ,pos,val);
 86  
87      // update max value 
88      tree[root].mmax = max(a,b);
 89  
90      return tree[root].mmax;
 91  }
 92  
93  
94  // int query(int root,int s,int e)
 95  // {
 96  //      if(tree[root].l == s && tree[root].r == e)
 97  //         return tree[root].mmax;
 98 //     if(e <= tree[root].mid())
 99 //         return query(root*2,s,e);
100 //     else
101 //         if(s > tree[root].mid())
102 //             return query(root*2+1,s,e);
103 //         else
104 //             return max(query(root*2,s,tree[root].mid()),query(root*2+1,tree[root].mid()+1,e));
105 // }
106 
107 int main()
108 {
109     int n,m,i,a,b;
110     char s;
111     while(scanf("%d %d",&n,&m)!=EOF)
112     {
113         for(i = 1;i <= n;i++)
114             scanf("%d",&num[i]);
115         BuildTree(1,1,n);
116         for(i = 1;i <= m;i++)
117         {
118             getchar();
119             scanf("%c",&s);
120             if(s == 'Q')
121             {
122                 scanf("%d %d",&a,&b);
123                 printf("%d\n",find(1,a,b));
124             }
125             if(s == 'U')
126             {
127                 scanf("%d %d",&a,&b);
128                 update(1,a,b);
129             }
130     }
131     }
132     return 0;
133 }