给定一个大小为N的数组arr[]和一个二维数组Q[][] ，由以下两种类型的查询组成：
- 1 X Val：更新arr[X] = Val 。
- 2 LR：求[L, R]范围内具有交替符号的数组元素的总和。
例子：
Input: arr[] = { 1, 2, 3, 4 }, Q[][] = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } }
Output:-2 2
Explanation:
Query1: Print (arr[0] – arr[1] + arr[2] – arr[3]) = 1 – 2 + 3 – 4 = -2
Query2: Updating arr[1] to 5 modifies arr[] to { 1, 5, 3, 4 }
Query3: Print (arr[1] – arr[2]) = 5 – 3 = 2
Input: arr[] = { 4, 2, 6, 1, 8, 9, 2}, Q = { { 2, 1, 4 }, { 1, 3, 4 }, { 2, 0, 3 } }
Output: -11 5
方法：这个问题可以使用Segment Tree来解决。这个想法是遍历数组并检查数组元素的索引是否为负，然后将数组元素乘以-1 。请按照以下步骤解决问题：
- 使用变量i遍历数组arr[]并检查索引i是否为奇数。如果发现为真，则更新arr[i] = -Val 。
- 对于类型1, X, Val 的查询，检查X是否为偶数。如果发现为真，则使用段树更新arr[X] = Val 。
- 否则，使用段树更新arr[X] = -Val 。
- 对于类型2 LR: 的查询，检查L是否为偶数。如果发现为真，则使用线段树打印[L, R]范围内数组元素的总和。
- 否则，使用线段树找到[L, R]范围内数组元素的总和，并打印得到的总和乘以-1 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to build the segment tree
void build(int tree[], int arr[], int start,
int end, int index)
{
// If current node is a leaf node
// of the segment tree
if (start == end) {
if (start % 2 == 0) {
// Update tree[index]
tree[index] = arr[start];
}
else {
// Update tree[index]
tree[index] = -arr[start];
}
return;
}
// Divide the segment tree
int mid = start + (end - start) / 2;
// Update on L segment tree
build(tree, arr, start, mid,
2 * index + 1);
// Update on R segment tree
build(tree, arr, mid + 1, end,
2 * index + 2);
// Find the sum from L subtree
// and R subtree
tree[index] = tree[2 * index + 1] + tree[2 * index + 2];
}
// Function to update elements at index pos
// by val in the segment tree
void update(int tree[], int index, int start,
int end, int pos, int val)
{
// If current node is a leaf node
if (start == end) {
// If current index is even
if (start % 2 == 0) {
// Update tree[index]
tree[index] = val;
}
else {
// Update tree[index]
tree[index] = -val;
}
return;
}
// Divide the segment tree elements
// into L and R subtree
int mid = start + (end - start) / 2;
// If element lies in L subtree
if (mid >= pos) {
update(tree, 2 * index + 1, start,
mid, pos, val);
}
else {
update(tree, 2 * index + 2, mid + 1,
end, pos, val);
}
// Update tree[index]
tree[index]
= tree[2 * index + 1] + tree[2 * index + 2];
}
// Function to find the sum of array elements
// in the range [L, R]
int FindSum(int tree[], int start, int end,
int L, int R, int index)
{
// If start and end not lies in
// the range [L, R]
if (L > end || R < start) {
return 0;
}
// If start and end comleately lies
// in the range [L, R]
if (L <= start && R >= end) {
return tree[index];
}
int mid = start + (end - start) / 2;
// Stores sum from left subtree
int X = FindSum(tree, start, mid, L,
R, 2 * index + 1);
// Stores sum from right subtree
int Y = FindSum(tree, mid + 1, end, L,
R, 2 * index + 2);
return X + Y;
}
int main()
{
int arr[] = { 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
int tree[4 * N + 5] = { 0 };
build(tree, arr, 0, N - 1, 0);
int Q[][3] = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } };
int cntQuey = 3;
for (int i = 0; i < cntQuey; i++) {
if (Q[i][0] == 1) {
update(tree, 0, 0, N - 1,
Q[i][1], Q[i][2]);
}
else {
if (Q[i][1] % 2 == 0) {
cout << FindSum(tree, 0, N - 1,
Q[i][1], Q[i][2], 0)
<< " ";
}
else {
cout << -FindSum(tree, 0, N - 1,
Q[i][1], Q[i][2], 0)
<< " ";
}
}
}
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to build the segment tree
static void build(int tree[], int arr[], int start,
int end, int index)
{
// If current node is a leaf node
// of the segment tree
if (start == end) {
if (start % 2 == 0) {
// Update tree[index]
tree[index] = arr[start];
}
else {
// Update tree[index]
tree[index] = -arr[start];
}
return;
}
// Divide the segment tree
int mid = start + (end - start) / 2;
// Update on L segment tree
build(tree, arr, start, mid,
2 * index + 1);
// Update on R segment tree
build(tree, arr, mid + 1, end,
2 * index + 2);
// Find the sum from L subtree
// and R subtree
tree[index] = tree[2 * index + 1] + tree[2 * index + 2];
}
// Function to update elements at index pos
// by val in the segment tree
static void update(int tree[], int index, int start,
int end, int pos, int val)
{
// If current node is a leaf node
if (start == end) {
// If current index is even
if (start % 2 == 0) {
// Update tree[index]
tree[index] = val;
}
else {
// Update tree[index]
tree[index] = -val;
}
return;
}
// Divide the segment tree elements
// into L and R subtree
int mid = start + (end - start) / 2;
// If element lies in L subtree
if (mid >= pos)
{
update(tree, 2 * index + 1, start,
mid, pos, val);
}
else
{
update(tree, 2 * index + 2, mid + 1,
end, pos, val);
}
// Update tree[index]
tree[index]
= tree[2 * index + 1] + tree[2 * index + 2];
}
// Function to find the sum of array elements
// in the range [L, R]
static int FindSum(int tree[], int start, int end,
int L, int R, int index)
{
// If start and end not lies in
// the range [L, R]
if (L > end || R < start)
{
return 0;
}
// If start and end comleately lies
// in the range [L, R]
if (L <= start && R >= end)
{
return tree[index];
}
int mid = start + (end - start) / 2;
// Stores sum from left subtree
int X = FindSum(tree, start, mid, L,
R, 2 * index + 1);
// Stores sum from right subtree
int Y = FindSum(tree, mid + 1, end, L,
R, 2 * index + 2);
return X + Y;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4 };
int N = arr.length;
int tree[] = new int[4 * N + 5];
Arrays.fill(tree, 0);
build(tree, arr, 0, N - 1, 0);
int Q[][] = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } };
int cntQuey = 3;
for (int i = 0; i < cntQuey; i++)
{
if (Q[i][0] == 1)
{
update(tree, 0, 0, N - 1,
Q[i][1], Q[i][2]);
}
else {
if (Q[i][1] % 2 == 0)
{
System.out.print(FindSum(tree, 0, N - 1,
Q[i][1], Q[i][2], 0) + " ");
}
else
{
System.out.print(-FindSum(tree, 0, N - 1,
Q[i][1], Q[i][2], 0)+ " ");
}
}
}
}
}
// This code is contributed by code_hunt.Python3
# Python3 program to implement
# the above approach
# Function to build the segment tree
def build(tree, arr, start, end, index):
# If current node is a leaf node
# of the segment tree
if (start == end):
if (start % 2 == 0):
# Update tree[index]
tree[index] = arr[start]
else:
# Update tree[index]
tree[index] = -arr[start]
return
# Divide the segment tree
mid = start + (end - start) // 2
# Update on L segment tree
build(tree, arr, start, mid, 2 * index + 1)
# Update on R segment tree
build(tree, arr, mid + 1, end, 2 * index + 2)
# Find the sum from L subtree
# and R subtree
tree[index] = tree[2 * index + 1] + tree[2 * index + 2]
# Function to update elements at index pos
# by val in the segment tree
def update(tree, index, start, end, pos, val):
# If current node is a leaf node
if (start == end):
# If current index is even
if (start % 2 == 0):
# Update tree[index]
tree[index] = val
else:
# Update tree[index]
tree[index] = -val
return
# Divide the segment tree elements
# into L and R subtree
mid = start + (end - start) // 2
# If element lies in L subtree
if (mid >= pos):
update(tree, 2 * index + 1, start, mid, pos, val)
else:
update(tree, 2 * index + 2, mid + 1, end, pos, val)
# Update tree[index]
tree[index] = tree[2 * index + 1] + tree[2 * index + 2]
# Function to find the sum of array elements
# in the range [L, R]
def FindSum(tree, start, end, L, R, index):
# If start and end not lies in
# the range [L, R]
if (L > end or R < start):
return 0
#If start and end comleately lies
#in the range [L, R]
if (L <= start and R >= end):
return tree[index]
mid = start + (end - start) // 2
# Stores sum from left subtree
X = FindSum(tree, start, mid, L, R, 2 * index + 1)
# Stores sum from right subtree
Y = FindSum(tree, mid + 1, end, L, R, 2 * index + 2)
return X + Y
# Driver code
if __name__ == '__main__':
arr = [1, 2, 3, 4]
N = len(arr)
tree = [0 for i in range(4 * N + 5)]
build(tree, arr, 0, N - 1, 0)
Q = [ [ 2, 0, 3 ], [ 1, 1, 5 ], [ 2, 1, 2 ] ]
cntQuey = 3
for i in range(cntQuey):
if (Q[i][0] == 1):
update(tree, 0, 0, N - 1, Q[i][1], Q[i][2])
else:
if (Q[i][1] % 2 == 0):
print(FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0),end=" ")
else:
print(-FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0),end=" ")
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to build the segment tree
static void build(int[] tree, int[] arr, int start,
int end, int index)
{
// If current node is a leaf node
// of the segment tree
if (start == end)
{
if (start % 2 == 0)
{
// Update tree[index]
tree[index] = arr[start];
}
else
{
// Update tree[index]
tree[index] = -arr[start];
}
return;
}
// Divide the segment tree
int mid = start + (end - start) / 2;
// Update on L segment tree
build(tree, arr, start, mid,
2 * index + 1);
// Update on R segment tree
build(tree, arr, mid + 1, end,
2 * index + 2);
// Find the sum from L subtree
// and R subtree
tree[index] = tree[2 * index + 1] + tree[2 * index + 2];
}
// Function to update elements at index pos
// by val in the segment tree
static void update(int[] tree, int index, int start,
int end, int pos, int val)
{
// If current node is a leaf node
if (start == end)
{
// If current index is even
if (start % 2 == 0)
{
// Update tree[index]
tree[index] = val;
}
else
{
// Update tree[index]
tree[index] = -val;
}
return;
}
// Divide the segment tree elements
// into L and R subtree
int mid = start + (end - start) / 2;
// If element lies in L subtree
if (mid >= pos)
{
update(tree, 2 * index + 1, start,
mid, pos, val);
}
else
{
update(tree, 2 * index + 2, mid + 1,
end, pos, val);
}
// Update tree[index]
tree[index]
= tree[2 * index + 1] + tree[2 * index + 2];
}
// Function to find the sum of array elements
// in the range [L, R]
static int FindSum(int[] tree, int start, int end,
int L, int R, int index)
{
// If start and end not lies in
// the range [L, R]
if (L > end || R < start)
{
return 0;
}
// If start and end comleately lies
// in the range [L, R]
if (L <= start && R >= end)
{
return tree[index];
}
int mid = start + (end - start) / 2;
// Stores sum from left subtree
int X = FindSum(tree, start, mid, L,
R, 2 * index + 1);
// Stores sum from right subtree
int Y = FindSum(tree, mid + 1, end, L,
R, 2 * index + 2);
return X + Y;
}
// Driver code
static void Main()
{
int[] arr = { 1, 2, 3, 4 };
int N = arr.Length;
int[] tree = new int[4 * N + 5];
build(tree, arr, 0, N - 1, 0);
int[,] Q = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } };
int cntQuey = 3;
for (int i = 0; i < cntQuey; i++)
{
if (Q[i, 0] == 1)
{
update(tree, 0, 0, N - 1,
Q[i, 1], Q[i, 2]);
}
else
{
if (Q[i, 1] % 2 == 0)
{
Console.Write(FindSum(tree, 0, N - 1,
Q[i, 1], Q[i, 2], 0) + " ");
}
else
{
Console.Write(-FindSum(tree, 0, N - 1,
Q[i, 1], Q[i, 2], 0) + " ");
}
}
}
}
}
// This code is contributed by divyesh072019.Javascript
输出：
-2 2时间复杂度： O(|Q| * log(N))
辅助空间： O(N)
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