Ruta con valor medio máximo
Dada una matriz cuadrada de tamaño N*N donde cada celda está asociada a un costo específico. Una ruta se define como una secuencia específica de celdas que comienza en la celda superior izquierda, se mueve solo hacia la derecha o hacia abajo y termina en la celda inferior derecha. Queremos encontrar un camino con el promedio máximo sobre todos los caminos existentes. El promedio se calcula como el costo total dividido por la cantidad de celdas visitadas en la ruta.
Ejemplos:
Input : Matrix = [1 2 3
4 5 6
7 8 9]
Output : 5.8
Path with maximum average is 1 -> 4 -> 7 -> 8 -> 9
Sum of the path is 29 and average is 29/5 = 5.8
Una observación interesante es que los únicos movimientos permitidos son hacia abajo y hacia la derecha; necesitamos N-1 movimientos hacia abajo y N-1 movimientos hacia la derecha para llegar al destino (abajo a la derecha). Entonces, cualquier camino desde la esquina superior izquierda hasta la esquina inferior derecha requiere 2N - 1 celdas. En promedio valor, el denominador es fijo y solo necesitamos maximizar el numerador. Por lo tanto, básicamente necesitamos encontrar la ruta de suma máxima. Calcular la suma máxima de la ruta es un problema clásico de programación dinámica si dp[i][j] representa la suma máxima hasta la celda (i j) desde (0 0), entonces en cada celda (i j) actualizamos dp[i][j] como se muestra a continuación
for all i 1 <= i <= N
dp[i][0] = dp[i-1][0] + cost[i][0];
for all j 1 <= j <= N
dp[0][j] = dp[0][j-1] + cost[0][j];
otherwise
dp[i][j] = max(dp[i-1][j] dp[i][j-1]) + cost[i][j];
Una vez que obtengamos la suma máxima de todos los caminos, dividiremos esta suma por (2N - 1) y obtendremos nuestro promedio máximo.
Implementación:
C++Java//C/C++ program to find maximum average cost path #includeusing namespace std ; // Maximum number of rows and/or columns const int M = 100 ; // method returns maximum average of all path of // cost matrix double maxAverageOfPath ( int cost [ M ][ M ] int N ) { int dp [ N + 1 ][ N + 1 ]; dp [ 0 ][ 0 ] = cost [ 0 ][ 0 ]; /* Initialize first column of total cost(dp) array */ for ( int i = 1 ; i < N ; i ++ ) dp [ i ][ 0 ] = dp [ i -1 ][ 0 ] + cost [ i ][ 0 ]; /* Initialize first row of dp array */ for ( int j = 1 ; j < N ; j ++ ) dp [ 0 ][ j ] = dp [ 0 ][ j -1 ] + cost [ 0 ][ j ]; /* Construct rest of the dp array */ for ( int i = 1 ; i < N ; i ++ ) for ( int j = 1 ; j <= N ; j ++ ) dp [ i ][ j ] = max ( dp [ i -1 ][ j ] dp [ i ][ j -1 ]) + cost [ i ][ j ]; // divide maximum sum by constant path // length : (2N - 1) for getting average return ( double ) dp [ N -1 ][ N -1 ] / ( 2 * N -1 ); } /* Driver program to test above functions */ int main () { int cost [ M ][ M ] = { { 1 2 3 } { 6 5 4 } { 7 3 9 } }; printf ( '%f' maxAverageOfPath ( cost 3 )); return 0 ; } C#// JAVA Code for Path with maximum average // value import java.io.* ; class GFG { // method returns maximum average of all // path of cost matrix public static double maxAverageOfPath ( int cost [][] int N ) { int dp [][] = new int [ N + 1 ][ N + 1 ] ; dp [ 0 ][ 0 ] = cost [ 0 ][ 0 ] ; /* Initialize first column of total cost(dp) array */ for ( int i = 1 ; i < N ; i ++ ) dp [ i ][ 0 ] = dp [ i - 1 ][ 0 ] + cost [ i ][ 0 ] ; /* Initialize first row of dp array */ for ( int j = 1 ; j < N ; j ++ ) dp [ 0 ][ j ] = dp [ 0 ][ j - 1 ] + cost [ 0 ][ j ] ; /* Construct rest of the dp array */ for ( int i = 1 ; i < N ; i ++ ) for ( int j = 1 ; j < N ; j ++ ) dp [ i ][ j ] = Math . max ( dp [ i - 1 ][ j ] dp [ i ][ j - 1 ] ) + cost [ i ][ j ] ; // divide maximum sum by constant path // length : (2N - 1) for getting average return ( double ) dp [ N - 1 ][ N - 1 ] / ( 2 * N - 1 ); } /* Driver program to test above function */ public static void main ( String [] args ) { int cost [][] = {{ 1 2 3 } { 6 5 4 } { 7 3 9 }}; System . out . println ( maxAverageOfPath ( cost 3 )); } } // This code is contributed by Arnav Kr. Mandal.JavaScript// C# Code for Path with maximum average // value using System ; class GFG { // method returns maximum average of all // path of cost matrix public static double maxAverageOfPath ( int [] cost int N ) { int [] dp = new int [ N + 1 N + 1 ]; dp [ 0 0 ] = cost [ 0 0 ]; /* Initialize first column of total cost(dp) array */ for ( int i = 1 ; i < N ; i ++ ) dp [ i 0 ] = dp [ i - 1 0 ] + cost [ i 0 ]; /* Initialize first row of dp array */ for ( int j = 1 ; j < N ; j ++ ) dp [ 0 j ] = dp [ 0 j - 1 ] + cost [ 0 j ]; /* Construct rest of the dp array */ for ( int i = 1 ; i < N ; i ++ ) for ( int j = 1 ; j < N ; j ++ ) dp [ i j ] = Math . Max ( dp [ i - 1 j ] dp [ i j - 1 ]) + cost [ i j ]; // divide maximum sum by constant path // length : (2N - 1) for getting average return ( double ) dp [ N - 1 N - 1 ] / ( 2 * N - 1 ); } // Driver Code public static void Main () { int [] cost = {{ 1 2 3 } { 6 5 4 } { 7 3 9 }}; Console . Write ( maxAverageOfPath ( cost 3 )); } } // This code is contributed by nitin mittal.PHP< script > // JavaScript Code for Path with maximum average value // method returns maximum average of all // path of cost matrix function maxAverageOfPath ( cost N ) { let dp = new Array ( N + 1 ); for ( let i = 0 ; i < N + 1 ; i ++ ) { dp [ i ] = new Array ( N + 1 ); for ( let j = 0 ; j < N + 1 ; j ++ ) { dp [ i ][ j ] = 0 ; } } dp [ 0 ][ 0 ] = cost [ 0 ][ 0 ]; /* Initialize first column of total cost(dp) array */ for ( let i = 1 ; i < N ; i ++ ) dp [ i ][ 0 ] = dp [ i - 1 ][ 0 ] + cost [ i ][ 0 ]; /* Initialize first row of dp array */ for ( let j = 1 ; j < N ; j ++ ) dp [ 0 ][ j ] = dp [ 0 ][ j - 1 ] + cost [ 0 ][ j ]; /* Construct rest of the dp array */ for ( let i = 1 ; i < N ; i ++ ) for ( let j = 1 ; j < N ; j ++ ) dp [ i ][ j ] = Math . max ( dp [ i - 1 ][ j ] dp [ i ][ j - 1 ]) + cost [ i ][ j ]; // divide maximum sum by constant path // length : (2N - 1) for getting average return dp [ N - 1 ][ N - 1 ] / ( 2 * N - 1 ); } let cost = [[ 1 2 3 ] [ 6 5 4 ] [ 7 3 9 ]]; document . write ( maxAverageOfPath ( cost 3 )); < /script>Python3// Php program to find maximum average cost path // method returns maximum average of all path of // cost matrix function maxAverageOfPath ( $cost $N ) { $dp = array ( array ()) ; $dp [ 0 ][ 0 ] = $cost [ 0 ][ 0 ]; /* Initialize first column of total cost(dp) array */ for ( $i = 1 ; $i < $N ; $i ++ ) $dp [ $i ][ 0 ] = $dp [ $i - 1 ][ 0 ] + $cost [ $i ][ 0 ]; /* Initialize first row of dp array */ for ( $j = 1 ; $j < $N ; $j ++ ) $dp [ 0 ][ $j ] = $dp [ 0 ][ $j - 1 ] + $cost [ 0 ][ $j ]; /* Construct rest of the dp array */ for ( $i = 1 ; $i < $N ; $i ++ ) { for ( $j = 1 ; $j <= $N ; $j ++ ) $dp [ $i ][ $j ] = max ( $dp [ $i - 1 ][ $j ] $dp [ $i ][ $j - 1 ]) + $cost [ $i ][ $j ]; } // divide maximum sum by constant path // length : (2N - 1) for getting average return $dp [ $N - 1 ][ $N - 1 ] / ( 2 * $N - 1 ); } // Driver code $cost = array ( array ( 1 2 3 ) array ( 6 5 4 ) array ( 7 3 9 ) ) ; echo maxAverageOfPath ( $cost 3 ) ; // This code is contributed by Ryuga ?>
Producción5.200000Complejidad del tiempo : EN 2 ) para una entrada dada N
Espacio Auxiliar: EN 2 ) para la entrada dada N.Método - 2: sin utilizar espacio extra N*N
Podemos usar la matriz de costos de entrada como dp para almacenar la respuesta. De esta manera no necesitamos una matriz dp adicional ni ese espacio adicional.
Una observación es que los únicos movimientos permitidos son hacia abajo y hacia la derecha; necesitamos N-1 movimientos hacia abajo y N-1 movimientos hacia la derecha para llegar al destino (abajo a la derecha). Entonces, cualquier camino desde la esquina superior izquierda hasta la esquina inferior derecha requiere 2N - 1 celda. En promedio valor, el denominador es fijo y solo necesitamos maximizar el numerador. Por lo tanto, básicamente necesitamos encontrar la ruta de suma máxima. Calcular la suma máxima de la ruta es un problema clásico de programación dinámica y además no necesitamos ningún valor de costo anterior [i] [j] después de calcular dp [i] [j] para que podamos modificar el valor de costo [i] [j] de modo que no necesitemos espacio adicional para dp [i] [j].
for all i 1 <= i < N
cost[i][0] = cost[i-1][0] + cost[i][0];
for all j 1 <= j < N
cost[0][j] = cost[0][j-1] + cost[0][j];
otherwise
cost[i][j] = max(cost[i-1][j] cost[i][j-1]) + cost[i][j];
A continuación se muestra la implementación del enfoque anterior:
C++Java// C++ program to find maximum average cost path #includeusing namespace std ; // Method returns maximum average of all path of cost matrix double maxAverageOfPath ( vector < vector < int >> cost ) { int N = cost . size (); // Initialize first column of total cost array for ( int i = 1 ; i < N ; i ++ ) cost [ i ][ 0 ] = cost [ i ][ 0 ] + cost [ i - 1 ][ 0 ]; // Initialize first row of array for ( int j = 1 ; j < N ; j ++ ) cost [ 0 ][ j ] = cost [ 0 ][ j - 1 ] + cost [ 0 ][ j ]; // Construct rest of the array for ( int i = 1 ; i < N ; i ++ ) for ( int j = 1 ; j <= N ; j ++ ) cost [ i ][ j ] = max ( cost [ i - 1 ][ j ] cost [ i ][ j - 1 ]) + cost [ i ][ j ]; // divide maximum sum by constant path // length : (2N - 1) for getting average return ( double ) cost [ N - 1 ][ N - 1 ] / ( 2 * N - 1 ); } // Driver program int main () { vector < vector < int >> cost = {{ 1 2 3 } { 6 5 4 } { 7 3 9 } }; cout < < maxAverageOfPath ( cost ); return 0 ; } C#// Java program to find maximum average cost path import java.io.* ; class GFG { // Method returns maximum average of all path of cost // matrix static double maxAverageOfPath ( int [][] cost ) { int N = cost . length ; // Initialize first column of total cost array for ( int i = 1 ; i < N ; i ++ ) cost [ i ][ 0 ] = cost [ i ][ 0 ] + cost [ i - 1 ][ 0 ] ; // Initialize first row of array for ( int j = 1 ; j < N ; j ++ ) cost [ 0 ][ j ] = cost [ 0 ][ j - 1 ] + cost [ 0 ][ j ] ; // Construct rest of the array for ( int i = 1 ; i < N ; i ++ ) for ( int j = 1 ; j < N ; j ++ ) cost [ i ][ j ] = Math . max ( cost [ i - 1 ][ j ] cost [ i ][ j - 1 ] ) + cost [ i ][ j ] ; // divide maximum sum by constant path // length : (2N - 1) for getting average return ( double ) cost [ N - 1 ][ N - 1 ] / ( 2 * N - 1 ); } // Driver program public static void main ( String [] args ) { int [][] cost = { { 1 2 3 } { 6 5 4 } { 7 3 9 } }; System . out . println ( maxAverageOfPath ( cost )); } } // This code is contributed by karandeep1234JavaScript// C# program to find maximum average cost path using System ; class GFG { // Method returns maximum average of all path of cost // matrix static double maxAverageOfPath ( int [ ] cost ) { int N = cost . GetLength ( 0 ); // Initialize first column of total cost array for ( int i = 1 ; i < N ; i ++ ) cost [ i 0 ] = cost [ i 0 ] + cost [ i - 1 0 ]; // Initialize first row of array for ( int j = 1 ; j < N ; j ++ ) cost [ 0 j ] = cost [ 0 j - 1 ] + cost [ 0 j ]; // Construct rest of the array for ( int i = 1 ; i < N ; i ++ ) for ( int j = 1 ; j < N ; j ++ ) cost [ i j ] = Math . Max ( cost [ i - 1 j ] cost [ i j - 1 ]) + cost [ i j ]; // divide maximum sum by constant path // length : (2N - 1) for getting average return ( double ) cost [ N - 1 N - 1 ] / ( 2 * N - 1 ); } // Driver program static void Main ( string [] args ) { int [ ] cost = { { 1 2 3 } { 6 5 4 } { 7 3 9 } }; Console . WriteLine ( maxAverageOfPath ( cost )); } } // This code is contributed by karandeep1234Python3// Method returns maximum average of all path of cost matrix function maxAverageOfPath ( cost ) { let N = cost . length ; // Initialize first column of total cost array for ( let i = 1 ; i < N ; i ++ ) cost [ i ][ 0 ] = cost [ i ][ 0 ] + cost [ i - 1 ][ 0 ]; // Initialize first row of array for ( let j = 1 ; j < N ; j ++ ) cost [ 0 ][ j ] = cost [ 0 ][ j - 1 ] + cost [ 0 ][ j ]; // Construct rest of the array for ( let i = 1 ; i < N ; i ++ ) for ( let j = 1 ; j <= N ; j ++ ) cost [ i ][ j ] = Math . max ( cost [ i - 1 ][ j ] cost [ i ][ j - 1 ]) + cost [ i ][ j ]; // divide maximum sum by constant path // length : (2N - 1) for getting average return ( cost [ N - 1 ][ N - 1 ]) / ( 2.0 * N - 1 ); } // Driver program let cost = [[ 1 2 3 ] [ 6 5 4 ] [ 7 3 9 ]]; console . log ( maxAverageOfPath ( cost )) // This code is contributed by karandeep1234.
Producción5.2Complejidad del tiempo: O(N*N)
Espacio Auxiliar: O(1)
