找出一个范围内不同数字的计数
给定一个大小为 N 的数组，其中只包含从 0 到 63 的数字，然后你会被问及关于它的问题。 查询如下:
- 1 X Y，即将索引 X 处的元素更改为 Y
- 2 L R，即打印介于左和右之间的不同元素的计数
例:
Input:
N = 7
ar = {1, 2, 1, 3, 1, 2, 1}
Q = 5
{ {2, 1, 4},
{1, 4, 2},
{1, 5, 2},
{2, 4, 6},
{2, 1, 7} }
Output:
3
1
2
Input:
N = 15
ar = {4, 6, 3, 2, 2, 3, 6, 5, 5, 5, 4, 2, 1, 5, 1}
Q = 5
{ {1, 6, 5},
{1, 4, 2},
{2, 6, 14},
{2, 6, 8},
{2, 1, 6} };
Output:
5
2
5
- 问题中形成的位掩码将表示第 I 个数字的存在与否，我们将通过检查我们的位掩码的第 I 个位来发现，如果第 I 个位是 on，这意味着第 I 个元素存在，否则不存在。
- 构建一个经典的段树，它的每个节点将包含一个位掩码，用于解码由特定节点定义的特定段中不同元素的数量。
- 以自下而上的方式构建段树，因此可以通过对该节点的左右子节点的位掩码执行按位“或”运算来轻松计算段树的当前节点的位掩码。叶节点将被单独处理。更新也将以同样的方式进行。
以下是上述方法的实现:
C++
// C++ Program to find the distinct
// elements in a range
#include <bits/stdc++.h>
using namespace std;
// Function to perform queries in a range
long long query(int start, int end, int left, int right,
int node, long long seg[])
{
// No overlap
if (end < left || start > right) {
return 0;
}
// Totally Overlap
else if (start >= left && end <= right) {
return seg[node];
}
// Partial Overlap
else {
int mid = (start + end) / 2;
// Finding the Answer
// for the left Child
long long leftChild = query(start, mid, left,
right, 2 * node, seg);
// Finding the Answer
// for the right Child
long long rightChild = query(mid + 1, end, left,
right, 2 * node + 1, seg);
// Combining the BitMasks
return (leftChild | rightChild);
}
}
// Function to perform update operation
// in the Segment seg
void update(int left, int right, int index, int Value,
int node, int ar[], long long seg[])
{
if (left == right) {
ar[index] = Value;
// Forming the BitMask
seg[node] = (1LL << Value);
return;
}
int mid = (left + right) / 2;
if (index > mid) {
// Updating the left Child
update(mid + 1, right, index, Value, 2 * node + 1, ar, seg);
}
else {
// Updating the right Child
update(left, mid, index, Value, 2 * node, ar, seg);
}
// Updating the BitMask
seg[node] = (seg[2 * node] | seg[2 * node + 1]);
}
// Building the Segment Tree
void build(int left, int right, int node, int ar[],
long long seg[])
{
if (left == right) {
// Building the Initial BitMask
seg[node] = (1LL << ar[left]);
return;
}
int mid = (left + right) / 2;
// Building the left seg tree
build(left, mid, 2 * node, ar, seg);
// Building the right seg tree
build(mid + 1, right, 2 * node + 1, ar, seg);
// Forming the BitMask
seg[node] = (seg[2 * node] | seg[2 * node + 1]);
}
// Utility Function to answer the queries
void getDistinctCount(vector<vector<int> >& queries,
int ar[], long long seg[], int n)
{
for (int i = 0; i < queries.size(); i++) {
int op = queries[i][0];
if (op == 2) {
int l = queries[i][1], r = queries[i][2];
long long tempMask = query(0, n - 1, l - 1,
r - 1, 1, seg);
int countOfBits = 0;
// Counting the set bits which denote the
// distinct elements
for (int i = 63; i >= 0; i--) {
if (tempMask & (1LL << i)) {
countOfBits++;
}
}
cout << countOfBits << '\n';
}
else {
int index = queries[i][1];
int val = queries[i][2];
// Updating the value
update(0, n - 1, index - 1, val, 1, ar, seg);
}
}
}
// Driver Code
int main()
{
int n = 7;
int ar[] = { 1, 2, 1, 3, 1, 2, 1 };
long long seg[4 * n] = { 0 };
build(0, n - 1, 1, ar, seg);
int q = 5;
vector<vector<int> > queries = {
{ 2, 1, 4 },
{ 1, 4, 2 },
{ 1, 5, 2 },
{ 2, 4, 6 },
{ 2, 1, 7 }
};
getDistinctCount(queries, ar, seg, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to find the distinct
// elements in a range
import java.util.*;
class GFG{
// Function to perform queries in a range
static long query(int start, int end, int left, int right,
int node, long seg[])
{
// No overlap
if (end < left || start > right) {
return 0;
}
// Totally Overlap
else if (start >= left && end <= right) {
return seg[node];
}
// Partial Overlap
else {
int mid = (start + end) / 2;
// Finding the Answer
// for the left Child
long leftChild = query(start, mid, left,
right, 2 * node, seg);
// Finding the Answer
// for the right Child
long rightChild = query(mid + 1, end, left,
right, 2 * node + 1, seg);
// Combining the BitMasks
return (leftChild | rightChild);
}
}
// Function to perform update operation
// in the Segment seg
static void update(int left, int right, int index, int Value,
int node, int ar[], long seg[])
{
if (left == right) {
ar[index] = Value;
// Forming the BitMask
seg[node] = (1L << Value);
return;
}
int mid = (left + right) / 2;
if (index > mid) {
// Updating the left Child
update(mid + 1, right, index, Value, 2 * node + 1, ar, seg);
}
else {
// Updating the right Child
update(left, mid, index, Value, 2 * node, ar, seg);
}
// Updating the BitMask
seg[node] = (seg[2 * node] | seg[2 * node + 1]);
}
// Building the Segment Tree
static void build(int left, int right, int node, int ar[],
long seg[])
{
if (left == right) {
// Building the Initial BitMask
seg[node] = (1L << ar[left]);
return;
}
int mid = (left + right) / 2;
// Building the left seg tree
build(left, mid, 2 * node, ar, seg);
// Building the right seg tree
build(mid + 1, right, 2 * node + 1, ar, seg);
// Forming the BitMask
seg[node] = (seg[2 * node] | seg[2 * node + 1]);
}
// Utility Function to answer the queries
static void getDistinctCount(int [][]queries,
int ar[], long seg[], int n)
{
for (int i = 0; i < queries.length; i++) {
int op = queries[i][0];
if (op == 2) {
int l = queries[i][1], r = queries[i][2];
long tempMask = query(0, n - 1, l - 1,
r - 1, 1, seg);
int countOfBits = 0;
// Counting the set bits which denote the
// distinct elements
for (int s = 63; s >= 0; s--) {
if ((tempMask & (1L << s))>0) {
countOfBits++;
}
}
System.out.println(countOfBits);
}
else {
int index = queries[i][1];
int val = queries[i][2];
// Updating the value
update(0, n - 1, index - 1, val, 1, ar, seg);
}
}
}
// Driver Code
public static void main(String[] args)
{
int n = 7;
int ar[] = { 1, 2, 1, 3, 1, 2, 1 };
long seg[] = new long[4 * n];
build(0, n - 1, 1, ar, seg);
int [][]queries = {
{ 2, 1, 4 },
{ 1, 4, 2 },
{ 1, 5, 2 },
{ 2, 4, 6 },
{ 2, 1, 7 }
};
getDistinctCount(queries, ar, seg, n);
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python3 Program to find the distinct
# elements in a range
# Function to perform queries in a range
def query(start, end, left,
right, node, seg):
# No overlap
if (end < left or start > right):
return 0
# Totally Overlap
elif (start >= left and
end <= right):
return seg[node]
# Partial Overlap
else:
mid = (start + end) // 2
# Finding the Answer
# for the left Child
leftChild = query(start, mid, left,
right, 2 * node, seg)
# Finding the Answer
# for the right Child
rightChild = query(mid + 1, end, left,
right, 2 * node + 1, seg)
# Combining the BitMasks
return (leftChild | rightChild)
# Function to perform update operation
# in the Segment seg
def update(left, right, index,
Value, node, ar, seg):
if (left == right):
ar[index] = Value
# Forming the BitMask
seg[node] = (1 << Value)
return
mid = (left + right) // 2
if (index > mid):
# Updating the left Child
update(mid + 1, right, index,
Value, 2 * node + 1, ar, seg)
else:
# Updating the right Child
update(left, mid, index,
Value, 2 * node, ar, seg)
# Updating the BitMask
seg[node] = (seg[2 * node] |
seg[2 * node + 1])
# Building the Segment Tree
def build(left, right, node, ar, eg):
if (left == right):
# Building the Initial BitMask
seg[node] = (1 << ar[left])
return
mid = (left + right) // 2
# Building the left seg tree
build(left, mid, 2 * node, ar, seg)
# Building the right seg tree
build(mid + 1, right,
2 * node + 1, ar, seg)
# Forming the BitMask
seg[node] = (seg[2 * node] |
seg[2 * node + 1])
# Utility Function to answer the queries
def getDistinctCount(queries, ar, seg, n):
for i in range(len(queries)):
op = queries[i][0]
if (op == 2):
l = queries[i][1]
r = queries[i][2]
tempMask = query(0, n - 1, l - 1,
r - 1, 1, seg)
countOfBits = 0
# Counting the set bits which denote
# the distinct elements
for i in range(63, -1, -1):
if (tempMask & (1 << i)):
countOfBits += 1
print(countOfBits)
else:
index = queries[i][1]
val = queries[i][2]
# Updating the value
update(0, n - 1, index - 1,
val, 1, ar, seg)
# Driver Code
if __name__ == '__main__':
n = 7
ar = [1, 2, 1, 3, 1, 2, 1]
seg = [0] * 4 * n
build(0, n - 1, 1, ar, seg)
q = 5
queries = [[ 2, 1, 4 ],
[ 1, 4, 2 ],
[ 1, 5, 2 ],
[ 2, 4, 6 ],
[ 2, 1, 7 ]]
getDistinctCount(queries, ar, seg, n)
# This code is contributed by Mohit Kumar
C
// C# Program to find the distinct
// elements in a range
using System;
class GFG{
// Function to perform queries in a range
static long query(int start, int end, int left, int right,
int node, long []seg)
{
// No overlap
if (end < left || start > right) {
return 0;
}
// Totally Overlap
else if (start >= left && end <= right) {
return seg[node];
}
// Partial Overlap
else {
int mid = (start + end) / 2;
// Finding the Answer
// for the left Child
long leftChild = query(start, mid, left,
right, 2 * node, seg);
// Finding the Answer
// for the right Child
long rightChild = query(mid + 1, end, left,
right, 2 * node + 1, seg);
// Combining the BitMasks
return (leftChild | rightChild);
}
}
// Function to perform update operation
// in the Segment seg
static void update(int left, int right, int index, int Value,
int node, int []ar, long []seg)
{
if (left == right) {
ar[index] = Value;
// Forming the BitMask
seg[node] = (1L << Value);
return;
}
int mid = (left + right) / 2;
if (index > mid) {
// Updating the left Child
update(mid + 1, right, index, Value, 2 * node + 1, ar, seg);
}
else {
// Updating the right Child
update(left, mid, index, Value, 2 * node, ar, seg);
}
// Updating the BitMask
seg[node] = (seg[2 * node] | seg[2 * node + 1]);
}
// Building the Segment Tree
static void build(int left, int right, int node, int []ar,
long []seg)
{
if (left == right) {
// Building the Initial BitMask
seg[node] = (1L << ar[left]);
return;
}
int mid = (left + right) / 2;
// Building the left seg tree
build(left, mid, 2 * node, ar, seg);
// Building the right seg tree
build(mid + 1, right, 2 * node + 1, ar, seg);
// Forming the BitMask
seg[node] = (seg[2 * node] | seg[2 * node + 1]);
}
// Utility Function to answer the queries
static void getDistinctCount(int [,]queries,
int []ar, long []seg, int n)
{
for (int i = 0; i < queries.GetLength(0); i++) {
int op = queries[i,0];
if (op == 2) {
int l = queries[i,1], r = queries[i,2];
long tempMask = query(0, n - 1, l - 1,
r - 1, 1, seg);
int countOfBits = 0;
// Counting the set bits which denote the
// distinct elements
for (int s = 63; s >= 0; s--) {
if ((tempMask & (1L << s))>0) {
countOfBits++;
}
}
Console.WriteLine(countOfBits);
}
else {
int index = queries[i,1];
int val = queries[i,2];
// Updating the value
update(0, n - 1, index - 1, val, 1, ar, seg);
}
}
}
// Driver Code
public static void Main(String[] args)
{
int n = 7;
int []ar = { 1, 2, 1, 3, 1, 2, 1 };
long []seg = new long[4 * n];
build(0, n - 1, 1, ar, seg);
int [,]queries = {
{ 2, 1, 4 },
{ 1, 4, 2 },
{ 1, 5, 2 },
{ 2, 4, 6 },
{ 2, 1, 7 }
};
getDistinctCount(queries, ar, seg, n);
}
}
// This code is contributed by 29AjayKumar
Output:
3
1
2
时间复杂度: O(N + Q*Log(N))
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