Dadas dos arrays R[] y C[] que constan de N enteros tales que se puede formar una array de dimensiones N x N cuyo elemento en el índice (i, j) viene dado por (R[i] + C[j]) para todos los pares posibles de las dos arrays. Dada una array Consultas[][] en la que cada fila representa una consulta del tipo {A, B, X, Y} , la tarea de cada consulta es verificar si existe una ruta desde la celda (A, B) a ( X, Y) en la array que consta solo de números pares o no. Si es cierto, escriba » Sí» . De lo contrario, imprima “No” .
Nota: Para una ruta desde la celda actual (i, j), los únicos movimientos posibles son (i + 1, j) , (i, j + 1) , (i – 1, j) y (i, j – 1) .
Ejemplos:
Entrada: R[] = {6, 2, 7, 8, 3}, C[] = {3, 4, 8, 5, 1}, Consultas[][] = {{2 2 1 3}, {4 2 4 3}, {5 1 3 4}}
Salida: Si Si No
Explicación:
La array formada es:
{{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}}
Consulta1: Existe una ruta con número par desde la celda (2, 2) a la (1) , 3) con todas las celdas como número par. El camino es (2, 2) -> (2, 3) -> (1, 3).
Consulta 2: existe una ruta con número par desde la celda (4, 2) a (4, 3) con todas las celdas como número par. El camino es (4, 2) -> (4, 3).
Consulta 3: No existe ninguna ruta con todos los elementos parejos.Entrada: R[] = {1, 2, 4, 8, 5}, C[] = {2, 5, 9, 7, 6}, Consultas[][] = {{2 2 1 3}, {1 4 1 3}, {5 1 3 4}}
Salida: No Si No
Explicación:
La array formada es:
{{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}}
Consulta 1: No existe una ruta con todos los elementos pares.
Consulta 2: existe una ruta con número par desde la celda (1, 4) a (1, 3) con todas las celdas como número par. El camino es (1, 4) -> (1, 3).
Consulta 3: No existe ninguna ruta con todos los elementos parejos.
Enfoque: la idea es precalcular la suma del prefijo de las arrays R[] y C[] y luego verificar la ruta válida con las condiciones dadas. A continuación se muestran los pasos:
- La idea es iterar desde la celda (r, c) a (r + 1, c) solo si la paridad de R[r] y R[r + 1] es la misma porque solo cambia un elemento.
- De manera similar, para las columnas de (r, c) a (r, c + 1) , la paridad de C y C debe ser la misma.
- Realice las operaciones anteriores para (r – 1, c) y (r, c – 1) .
- Ahora, para cada consulta, procede comprobando si todos los elementos de R[] desde r = (min(A, B) hasta max(A, B)) tienen la misma paridad y si todos los elementos de C[] desde c = (min (X, Y) a max(X, Y)) tienen la misma paridad. Esto se puede verificar en tiempo O (1) por consulta precalculando la suma de paridad de elementos del prefijo.
- Si la condición anterior se cumple para todas las celdas, imprima «Sí» . De lo contrario, imprima «No» .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find whether the path
// exists with all element even or not
void answerQueries(int n, int q,
vector<int>& a,
vector<int>& b,
vector<vector<int> >& 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<int> R{ 6, 2, 7, 8, 3 };
vector<int> C{ 3, 4, 8, 5, 1 };
// Given queries[][]
vector<vector<int> > 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 bikram2001jha
Python3
# 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 shikhasingrajput
C#
// 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
Javascript
<script>
// Javascript program to implement
// the above approach
// Function to find whether the path
// exists with all element even or not
function 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(let 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(let 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(let i = 0; i < q; i++)
{
let ra = Queries[i][0];
let ca = Queries[i][1];
let rb = Queries[i][2];
let cb = Queries[i][3];
let r2 = Math.max(ra, rb),
r1 = Math.min(ra, rb);
let 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
document.write("Yes" + " ");
continue;
}
}
// No path exists
document.write("No" + " ");
}
}
// Driver Code
// Given N, Q
let N = 5, Q = 3;
// Given array R[] and C[]
let R = [0, 6, 2, 7, 8, 3 ];
let C = [0, 3, 4, 8, 5, 1 ];
// Given queries[][]
let 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 avijitmondal1998.
</script>
Producción
Yes Yes No
Complejidad temporal: O(N + Q)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por rohitpal210 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA