给定两个由N个整数组成的数组R []和C [] ，这样就可以形成尺寸为N x N的矩阵，其索引为(i，j)的元素由(R [i] + C [j])给出两个数组中所有可能的对。给定矩阵Queries [] [] ，每行代表一个{A，B，X，Y}类型的查询，每个查询的任务是检查是否存在从单元格(A，B)到( X，Y)是否仅由偶数组成。如果发现是真的，则打印“是” 。否则，打印“否” 。
注意：对于从当前单元格(i，j)出发的路径，唯一可能的移动是(i + 1，j) ， (i，j + 1) ， (i – 1，j)和(i，j – 1) 。
例子：
Input: R[] = {6, 2, 7, 8, 3}, C[] = {3, 4, 8, 5, 1}, Queries[][] = {{2 2 1 3}, {4 2 4 3}, {5 1 3 4}}
Output: Yes Yes No
Explanation:
The matrix formed is:
{{9, 10, 14, 11, 7}
{5, 6, 10, 7, 3}
{10, 11, 15, 12, 8}
{11, 12, 16, 13, 9}
{6, 7, 11, 8, 4}}
Query1: There exist a path with even number from cell (2, 2) to (1, 3) with all cell as even number. The path is (2, 2) -> (2, 3) -> (1, 3).
Query2: There exist a path with even number from cell (4, 2) to (4, 3) with all cell as even number. The path is (4, 2) -> (4, 3).
Query3: No path exists with all elements even.
Input: R[] = {1, 2, 4, 8, 5}, C[] = {2, 5, 9, 7, 6}, Queries[][] = {{2 2 1 3}, {1 4 1 3}, {5 1 3 4}}
Output: No Yes No
Explanation:
The matrix formed is:
{{3, 6, 10, 8, 7}
{4, 7, 11, 9, 8}
{6, 9, 13, 11, 10}
{10, 13, 17, 15, 14}
{7, 10, 14, 12, 11}}
Query1: No path exists with all elements even.
Query2: There exist a path with even number from cell (1, 4) to (1, 3) with all cell as even number. The path is (1, 4) -> (1, 3).
Query3: No path exists with all elements even.
方法：想法是预先计算数组R []和C []的前缀和，然后在给定条件下检查有效路径。步骤如下：
- 这个想法是仅在R [r]和R [r + 1]的奇偶性相同的情况下才从单元(r，c)迭代到(r + 1，c)，因为只有一个元素发生变化。
- 类似地，对于(r，c)到(r，c +1)的列， C和C的奇偶校验应该相同。
- 对(r – 1，c)和(r，c – 1)执行上述操作。
- 现在，对于每个查询，通过检查进入如果R []的从r =(分钟(A，B)到最大A，B)()的所有元素具有相同的奇偶性，并从C C []中的所有元素=(分(X，Y)到max(X，Y))具有相同的奇偶校验。可以通过预先计算元素的奇偶校验前缀总和来在每个查询的O(1)时间中对此进行检查。
- 如果以上条件对所有单元格均成立，则打印“是”，否则打印“否” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find whether the path
// exists with all element even or not
void answerQueries(int n, int q,
vector& a,
vector& b,
vector >& Queries)
{
// Add 0 to beginning to make
// 1-based-indexing
a.insert(a.begin(), 0);
b.insert(b.begin(), 0);
// Change elements based on parity
// and take prefix sum
// 1 is odd and 0 is even
// Traverse the array R[]
for (int i = 1; i <= n; i++) {
// Taking & to check parity
if (a[i] & 1)
a[i] = 1;
else
a[i] = 0;
// Build prefix sum array
a[i] += a[i - 1];
}
// Traverse the array C[]
for (int i = 1; i <= n; i++) {
// Taking & to check parity
if (b[i] & 1)
b[i] = 1;
else
b[i] = 0;
// Build prefix sum array
b[i] += b[i - 1];
}
// Traverse the matrix Queries[][]
for (int i = 0; i < q; i++) {
int ra = Queries[i][0];
int ca = Queries[i][1];
int rb = Queries[i][2];
int cb = Queries[i][3];
int r2 = max(ra, rb), r1 = min(ra, rb);
int c2 = max(ca, cb), c1 = min(ca, cb);
// Check if all numbers are of
// same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0)
|| (a[r2] - a[r1 - 1]
== r2 - r1 + 1)) {
// Check if all numbers are of
// same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0)
|| (b[c2] - b[c1 - 1]
== c2 - c1 + 1)) {
// Path exists
cout << "Yes" << ' ';
continue;
}
}
// No path exists
cout << "No" << ' ';
}
}
// Driver Code
int main()
{
// Given N, Q
int N = 5, Q = 3;
// Given array R[] and C[]
vector R{ 6, 2, 7, 8, 3 };
vector C{ 3, 4, 8, 5, 1 };
// Given queries[][]
vector > Queries{ { 2, 2, 1, 3 },
{ 4, 2, 4, 3 },
{ 5, 1, 3, 4 } };
// Function Call
answerQueries(N, Q, R, C, Queries);
return 0;
} Java
// Java program for the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Function to find whether the path
// exists with all element even or not
static void answerQueries(int n, int q,
int[] a, int[] b,
int[][] Queries)
{
// Add 0 to beginning to make
// 1-based-indexing
// a[0]=0;
// b[0]=0;
// Change elements based on parity
// and take prefix sum
// 1 is odd and 0 is even
// Traverse the array R[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((a[i] & 1) != 0)
a[i] = 1;
else
a[i] = 0;
// Build prefix sum array
a[i] += a[i - 1];
}
// Traverse the array C[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((b[i] & 1) != 0)
b[i] = 1;
else
b[i] = 0;
// Build prefix sum array
b[i] += b[i - 1];
}
// Traverse the matrix Queries[][]
for(int i = 0; i < q; i++)
{
int ra = Queries[i][0];
int ca = Queries[i][1];
int rb = Queries[i][2];
int cb = Queries[i][3];
int r2 = Math.max(ra, rb),
r1 = Math.min(ra, rb);
int c2 = Math.max(ca, cb),
c1 = Math.min(ca, cb);
// Check if all numbers are of
// same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0) ||
(a[r2] - a[r1 - 1] == r2 - r1 + 1))
{
// Check if all numbers are of
// same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0) ||
(b[c2] - b[c1 - 1] == c2 - c1 + 1))
{
// Path exists
System.out.print("Yes" + " ");
continue;
}
}
// No path exists
System.out.print("No" + " ");
}
}
// Driver Code
public static void main (String[] args)
throws java.lang.Exception
{
// Given N, Q
int N = 5, Q = 3;
// Given array R[] and C[]
int R[] = { 0, 6, 2, 7, 8, 3 };
int C[] = { 0, 3, 4, 8, 5, 1 };
// Given queries[][]
int[][] Queries = { { 2, 2, 1, 3 },
{ 4, 2, 4, 3 },
{ 5, 1, 3, 4 }};
// Function call
answerQueries(N, Q, R, C, Queries);
}
}
// This code is contributed by bikram2001jhaPython3
# Python3 program for
# the above approach
# Function to find whether the path
# exists with all element even or not
def answerQueries(n, q, a, b, Queries):
# Add 0 to beginning to make
# 1-based-indexing
# a[0]=0;
# b[0]=0;
# Change elements based on parity
# and take prefix sum
# 1 is odd and 0 is even
# Traverse the array R
for i in range(1, n + 1):
# Taking & to check parity
if ((a[i] & 1) != 0):
a[i] = 1;
else:
a[i] = 0;
# Build prefix sum array
a[i] += a[i - 1];
# Traverse the array C
for i in range(1, n + 1):
# Taking & to check parity
if ((b[i] & 1) != 0):
b[i] = 1;
else:
b[i] = 0;
# Build prefix sum array
b[i] += b[i - 1];
# Traverse the matrix Queries
for i in range(0, q):
ra = Queries[i][0];
ca = Queries[i][1];
rb = Queries[i][2];
cb = Queries[i][3];
r2 = max(ra, rb); r1 = min(ra, rb);
c2 = max(ca, cb); c1 = min(ca, cb);
# Check if all numbers are of
# same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0) or
(a[r2] - a[r1 - 1] == r2 - r1 + 1)):
# Check if all numbers are of
# same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0) or
(b[c2] - b[c1 - 1] == c2 - c1 + 1)):
# Path exists
print("Yes", end = " ");
continue;
# No path exists
print("No", end = " ");
# Driver Code
if __name__ == '__main__':
# Given N, Q
N = 5;
Q = 3;
# Given array R and C
R = [0, 6, 2, 7, 8, 3];
C = [0, 3, 4, 8, 5, 1];
# Given queries
Queries = [[2, 2, 1, 3],
[4, 2, 4, 3],
[5, 1, 3, 4]];
# Function call
answerQueries(N, Q, R, C, Queries);
# This code is contributed by shikhasingrajputC#
// C# program for
// the above approach
using System;
class GFG{
// Function to find whether the path
// exists with all element even or not
static void answerQueries(int n, int q,
int[] a, int[] b,
int[,] Queries)
{
// Add 0 to beginning to make
// 1-based-indexing
// a[0]=0;
// b[0]=0;
// Change elements based on parity
// and take prefix sum
// 1 is odd and 0 is even
// Traverse the array R[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((a[i] & 1) != 0)
a[i] = 1;
else
a[i] = 0;
// Build prefix sum array
a[i] += a[i - 1];
}
// Traverse the array C[]
for(int i = 1; i <= n; i++)
{
// Taking & to check parity
if ((b[i] & 1) != 0)
b[i] = 1;
else
b[i] = 0;
// Build prefix sum array
b[i] += b[i - 1];
}
// Traverse the matrix Queries[,]
for(int i = 0; i < q; i++)
{
int ra = Queries[i, 0];
int ca = Queries[i, 1];
int rb = Queries[i, 2];
int cb = Queries[i, 3];
int r2 = Math.Max(ra, rb),
r1 = Math.Min(ra, rb);
int c2 = Math.Max(ca, cb),
c1 = Math.Min(ca, cb);
// Check if all numbers are of
// same parity between (r1 to r2)
if ((a[r2] - a[r1 - 1] == 0) ||
(a[r2] - a[r1 - 1] == r2 - r1 + 1))
{
// Check if all numbers are of
// same parity between (r1 to r2)
if ((b[c2] - b[c1 - 1] == 0) ||
(b[c2] - b[c1 - 1] == c2 - c1 + 1))
{
// Path exists
Console.Write("Yes" + " ");
continue;
}
}
// No path exists
Console.Write("No" + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
// Given N, Q
int N = 5, Q = 3;
// Given array R[] and C[]
int []R = {0, 6, 2, 7, 8, 3};
int []C = {0, 3, 4, 8, 5, 1};
// Given [,]queries
int[,] Queries = {{2, 2, 1, 3},
{4, 2, 4, 3},
{5, 1, 3, 4}};
// Function call
answerQueries(N, Q, R, C, Queries);
}
}
// This code is contributed by Princi Singh输出
Yes Yes No 时间复杂度： O(N + Q)
辅助空间： O(1)