线段树的概念

Tushar的讲解视频Segment Tree Range Minimum Query清晰明了。

线段树的例子

Sum of given range

We have an array arr[0 . . . n-1]. We should be able to

  1. Find the sum of elements from index l to r where 0 <= l <= r <= n-1
  2. Change value of a specified element of the array to a new value x. We need to do arr[i] = x where 0 <= i <= n-1.

Representation of Segment trees

  1. Leaf Nodes are the elements of the input array.
  2. Each internal node represents some merging of the leaf nodes. The merging may be different for different problems. For this problem, merging is sum of leaves under a node.

An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i+1, right child at 2*i+2 and the parent is at(i-1)/2.

Construction of Segment Tree from given array
We start with a segment arr[0 . . . n-1]. and every time we divide the current segment into two halves(if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment, we store the sum in the corresponding node.
All levels of the constructed segment tree will be completely filled except the last level. Also, the tree will be a Full Binary Tree because we always divide segments in two halves at every level. Since the constructed tree is always a full binary tree with n leaves, there will be n-1 internal nodes. So total number of nodes will be 2*n – 1.

Height of the segment tree will be logN. Since the tree is represented using array and relation between parent and child indexes must be maintained, size of memory allocated for segment tree will be 2*pow(2, logN)-1.
Query for Sum of given range
Once the tree is constructed, how to get the sum using the constructed segment tree. Following is the algorithm to get the sum of elements.

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int getSum(node, l, r) {
   if the range of the node is within l and r
        return value in the node
   else if the range of the node is completely outside l and r
        return 0
   else
    return getSum(node's left child, l, r) + 
           getSum(node's right child, l, r)
}

Update a value
Like tree construction and query operations, the update can also be done recursively. We are given an index which needs to be updated. Let diff be the value to be added. We start from root of the segment tree and add diff to all nodes which have given index in their range. If a node doesn’t have given index in its range, we don’t make any changes to that node.
Implementation
Following is the implementation of segment tree. The program implements construction of segment tree for any given array. It also implements query and update operations.

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// CPP program to show segment tree operations like construction, query 
// and update 
#include <iostream> 
#include <cstdio>
#include <math.h> 

using namespace std;
  
// A utility function to get the middle index from corner indexes. 
int getMid(int s, int e) {  return s + (e -s)/2;  } 
  
/*  A recursive function to get the sum of values in given range 
    of the array. The following are parameters for this function. 
  
    st    --> Pointer to segment tree 
    si    --> Index of current node in the segment tree. Initially 
              0 is passed as root is always at index 0 
    ss & se  --> Starting and ending indexes of the segment represented 
                 by current node, i.e., st[si] 
    qs & qe  --> Starting and ending indexes of query range */
int getSumUtil(int *st, int ss, int se, int qs, int qe, int si) { 
    // If segment of this node is a part of given range, then return 
    // the sum of the segment 
    if (qs <= ss && qe >= se) 
        return st[si]; 
  
    // If segment of this node is outside the given range 
    if (se < qs || ss > qe) 
        return 0; 
  
    // If a part of this segment overlaps with the given range 
    int mid = getMid(ss, se); 
    return getSumUtil(st, ss, mid, qs, qe, 2*si+1) + 
           getSumUtil(st, mid+1, se, qs, qe, 2*si+2); 
} 
  
/* A recursive function to update the nodes which have the given  
   index in their range. The following are parameters 
    st, si, ss and se are same as getSumUtil() 
    i    --> index of the element to be updated. This index is  
             in the input array. 
   diff --> Value to be added to all nodes which have i in range */
void updateValueUtil(int *st, int ss, int se, int i, int diff, int si) { 
    // Base Case: If the input index lies outside the range of  
    // this segment 
    if (i < ss || i > se) 
        return; 
  
    // If the input index is in range of this node, then update  
    // the value of the node and its children 
    st[si] = st[si] + diff; 
    if (se != ss) { 
        int mid = getMid(ss, se); 
        updateValueUtil(st, ss, mid, i, diff, 2*si + 1); 
        updateValueUtil(st, mid+1, se, i, diff, 2*si + 2); 
    } 
} 
  
// The function to update a value in input array and segment tree. 
// It uses updateValueUtil() to update the value in segment tree 
void updateValue(int arr[], int *st, int n, int i, int new_val) { 
    // Check for erroneous input index 
    if (i < 0 || i > n-1) { 
        printf("Invalid Input"); 
        return; 
    } 
  
    // Get the difference between new value and old value 
    int diff = new_val - arr[i]; 
  
    // Update the value in array 
    arr[i] = new_val; 
  
    // Update the values of nodes in segment tree 
    updateValueUtil(st, 0, n-1, i, diff, 0); 
} 
  
// Return sum of elements in range from index qs (quey start) 
// to qe (query end).  It mainly uses getSumUtil() 
int getSum(int *st, int n, int qs, int qe) { 
    // Check for erroneous input values 
    if (qs < 0 || qe > n-1 || qs > qe) { 
        printf("Invalid Input"); 
        return -1; 
    } 
  
    return getSumUtil(st, 0, n-1, qs, qe, 0); 
} 
  
// A recursive function that constructs Segment Tree for array[ss..se]. 
// si is index of current node in segment tree st 
int constructSTUtil(int arr[], int ss, int se, int *st, int si) { 
    // If there is one element in array, store it in current node of 
    // segment tree and return 
    if (ss == se) { 
        st[si] = arr[ss]; 
        return arr[ss]; 
    } 
  
    // If there are more than one elements, then recur for left and 
    // right subtrees and store the sum of values in this node 
    int mid = getMid(ss, se); 
    st[si] =  constructSTUtil(arr, ss, mid, st, si*2+1) + 
              constructSTUtil(arr, mid+1, se, st, si*2+2); 
    return st[si]; 
} 
  
/* Function to construct segment tree from given array. This function 
   allocates memory for segment tree and calls constructSTUtil() to 
   fill the allocated memory */
int *constructST(int arr[], int n) { 
    // Allocate memory for the segment tree 
  
    //Height of segment tree 
    int x = (int)(ceil(log2(n)));  
  
    //Maximum size of segment tree 
    int max_size = 2*(int)pow(2, x) - 1;  
  
    // Allocate memory 
    int *st = new int[max_size]; 
  
    // Fill the allocated memory st 
    constructSTUtil(arr, 0, n-1, st, 0); 
  
    // Return the constructed segment tree 
    return st; 
} 
  
// Driver program to test above functions 
int main() { 
    int arr[] = {1, 3, 5, 7, 9, 11}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
  
    // Build segment tree from given array 
    int *st = constructST(arr, n); //确保n大于0
  
    // Print sum of values in array from index 1 to 3 
    cout<<"Sum of values in given range = "<<getSum(st, n, 1, 3)<<endl; 
             
  
    // Update: set arr[1] = 10 and update corresponding  
    // segment tree nodes 
    updateValue(arr, st, n, 1, 10); 
  
    // Find sum after the value is updated 
    cout<<"Updated sum of values in given range = "<<getSum(st, n, 1, 3)<<endl;
   
    return 0; 
}

Output:

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Sum of values in given range = 15
Updated sum of values in given range = 22

Time Complexity
Time Complexity for tree construction is O(n). There are total 2n-1 nodes, and value of every node is calculated only once in tree construction.

Time complexity to query is O(logN). To query a sum, we process at most four nodes at every level and number of levels is O(logN).

The time complexity of update is also O(logN). To update a leaf value, we process one node at every level and number of levels is O(logN).

Range Minimum Query

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#include <iostream> 
#include <cstdio>
#include <math.h> 
using namespace std;
  

int getMid(int s, int e) {  return s + (e -s)/2;  } 

int min(int a, int b) { return a >= b ? b : a; }
  
int getMinUtil(int *st, int ss, int se, int qs, int qe, int si) { 
    if (qs <= ss && qe >= se) 
        return st[si]; 
    if (se < qs || ss > qe) 
        return INT_MAX; 
    int mid = getMid(ss, se); 
    return min(getMinUtil(st, ss, mid, qs, qe, 2*si+1), getMinUtil(st, mid+1, se, qs, qe, 2*si+2)); 
} 
  

void updateValueUtil(int *st, int ss, int se, int i, int new_val, int si) { 
    if (i < ss || i > se) 
        return; 
    st[si] = min(st[si], new_val); 
    if (se != ss) { 
        int mid = getMid(ss, se); 
        updateValueUtil(st, ss, mid, i, new_val, 2*si + 1); 
        updateValueUtil(st, mid+1, se, i, new_val, 2*si + 2); 
    } 
} 
  

void updateValue(int arr[], int *st, int n, int i, int new_val) { 
    if (i < 0 || i > n-1) { 
        printf("Invalid Input"); 
        return; 
    } 
    arr[i] = new_val; 
    updateValueUtil(st, 0, n-1, i, new_val, 0); 
} 
  
int getMin(int *st, int n, int qs, int qe) { 
    if (qs < 0 || qe > n-1 || qs > qe) { 
        printf("Invalid Input"); 
        return -1; 
    } 
    return getMinUtil(st, 0, n-1, qs, qe, 0); 
} 
  
int constructSTUtil(int arr[], int ss, int se, int *st, int si) { 
    if (ss == se) { 
        st[si] = arr[ss]; 
        return arr[ss]; 
    } 
    int mid = getMid(ss, se); 
    st[si] =  min(constructSTUtil(arr, ss, mid, st, si*2+1), constructSTUtil(arr, mid+1, se, st, si*2+2)); 
    return st[si]; 
} 
  
int *constructST(int arr[], int n) { 
    int x = (int)(ceil(log2(n)));  
    int max_size = 2*(int)pow(2, x) - 1;  
    int *st = new int[max_size]; 
    constructSTUtil(arr, 0, n-1, st, 0); 
    return st; 
} 
  
int main() { 
    int arr[] = {1, 3, 2, 7, 9, 11}; 
    int n = sizeof(arr)/sizeof(arr[0]); 

    int *st = constructST(arr, n); //确保n大于0
    updateValue(arr, st, n, 1, -10); 
  
    cout<<"min of values in given range = "<<getMin(st, n, 1, 3)<<endl; 

    return 0; 
}

参考资料：

  1. Tushar Segment Tree Range Minimum Query
  2. Segment Tree | Set 1 (Sum of given range)
  3. Segment Tree | Set 2 (Range Minimum Query)