Ceļš ar maksimālo vidējo vērtību
Dota kvadrātveida matrica ar izmēru N*N, kur katra šūna ir saistīta ar konkrētām izmaksām. Ceļš tiek definēts kā noteikta šūnu secība, kas sākas no augšējās kreisās šūnas, pārvietojoties tikai pa labi vai uz leju, un beidzas apakšējā labajā šūnā. Mēs vēlamies atrast ceļu ar maksimālo vidējo vērtību visiem esošajiem ceļiem. Vidējā vērtība tiek aprēķināta, kopējās izmaksas dalītas ar ceļā apmeklēto šūnu skaitu.
Piemēri:
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
Viens interesants novērojums ir tas, ka vienīgās atļautās kustības ir uz leju, un pa labi, lai sasniegtu galamērķi, ir vajadzīgas N-1 kustības uz leju un N-1 pa labi. Tātad jebkuram ceļam no augšējā kreisā stūra uz apakšējo labo stūri ir vajadzīgas 2N–1 šūnas. In vidēji vērtība, saucējs ir fiksēts, un mums vienkārši jāpalielina skaitītājs. Tāpēc mums būtībā jāatrod maksimālās summas ceļš. Maksimālās ceļa summas aprēķināšana ir klasiska dinamiskas programmēšanas problēma, ja dp[i][j] apzīmē maksimālo summu līdz šūnai (i j) no (0 0), tad katrā šūnā (i j) mēs atjauninām dp[i][j], kā norādīts tālāk.
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];
Kad mēs iegūstam visu ceļu maksimālo summu, mēs šo summu dalīsim ar (2N - 1) un iegūsim maksimālo vidējo.
Īstenošana:
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 ?>
Izvade5.200000Laika sarežģītība : O(N 2 ) ievadei N
Palīgtelpa: O(N 2 ) dotajai ievadei N.2. metode: neizmantojot papildu N*N vietu
Mēs varam izmantot ievades izmaksu masīvu kā dp, lai saglabātu ans. tāpēc mums nav nepieciešams papildu dp masīvs vai nav nepieciešama papildu vieta.
Viens novērojums ir tāds, ka vienīgās atļautās kustības ir uz leju un pa labi, lai sasniegtu galamērķi, mums ir vajadzīgas N-1 kustības uz leju un N-1 pa labi, lai sasniegtu galamērķi (labajā apakšā). Tātad jebkuram ceļam no augšējā kreisā stūra uz apakšējo labo stūri ir nepieciešama 2N - 1 šūna. In vidēji vērtība, saucējs ir fiksēts, un mums vienkārši jāpalielina skaitītājs. Tāpēc mums būtībā jāatrod maksimālās summas ceļš. Maksimālās ceļa summas aprēķināšana ir klasiska dinamiskas programmēšanas problēma, turklāt mums nav nepieciešama iepriekšējā izmaksu [i][j] vērtība pēc dp[i][j] aprēķināšanas, lai mēs varētu mainīt izmaksu [i][j] vērtību tā, lai mums nebūtu nepieciešama papildu vieta 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];
Tālāk ir aprakstīta iepriekš minētās pieejas īstenošana.
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.
Izvade5.2Laika sarežģītība: O(N*N)
Palīgtelpa: O(1)
