Maksimalt areal av trekant med forskjellige toppunktfarger
Gitt en matrise av N rader og M kolonner består av tre verdier {r g b}. Oppgaven er å finne arealet av den største trekanten som har en side parallell med y-aksen, dvs. vertikal og fargen på alle tre toppunktene er forskjellige.
Eksempler:
Input : N = 4 M =5
mat[][] =
{
r r r r r
r r r r g
r r r r r
b b b b b
}
Output : 10
The maximum area of triangle is 10.
Triangle coordinates are (00) containing r (14) containing g (30) containing b.
Vi vet arealet av en trekant = 1/2 * base *høyde, så vi må maksimere basen og høyden til trekanten. Siden den ene siden er parallell med y-aksen, kan vi betrakte den siden som basis av trekanten.
C++
For å maksimere grunntallet kan vi finne den første og siste forekomsten av {r g b} for hver kolonne. Så vi har to sett med 3 verdier for hver kolonne. For base i en kolonne er ett toppunkt fra det første settet og det andre toppunktet fra det andre settet slik at de har forskjellige verdier.
For å maksimere høyden for en hvilken som helst kolonne som basis, må det tredje toppunktet velges slik at toppunktet skal være lengst fra kolonnen på venstre eller høyre side av kolonnen med en verdi som er forskjellig fra de to andre toppunktene.
Finn nå det maksimale arealet av trekanten for hver kolonne.
Nedenfor er implementeringen av denne tilnærmingen:
Java// C++ program to find maximum area of triangle // having different vertex color in a matrix. #includeusing namespace std ; #define R 4 #define C 5 // return the color value so that their corresponding // index can be access. int mapcolor ( char c ) { if ( c == 'r' ) return 0 ; else if ( c == 'g' ) return 1 ; else if ( c == 'b' ) return 2 ; } // Returns the maximum area of triangle from all // the possible triangles double findarea ( char mat [ R ][ C ] int r int c int top [ 3 ][ C ] int bottom [ 3 ][ C ] int left [ 3 ] int right [ 3 ]) { double ans = ( double ) 1 ; // for each column for ( int i = 0 ; i < c ; i ++ ) // for each top vertex for ( int x = 0 ; x < 3 ; x ++ ) // for each bottom vertex for ( int y = 0 ; y < 3 ; y ++ ) { // finding the third color of // vertex either on right or left. int z = 3 - x - y ; // finding area of triangle on left side of column. if ( x != y && top [ x ][ i ] != INT_MAX && bottom [ y ][ i ] != INT_MIN && left [ z ] != INT_MAX ) { ans = max ( ans (( double ) 1 / ( double ) 2 ) * ( bottom [ y ][ i ] - top [ x ][ i ]) * ( i - left [ z ])); } // finding area of triangle on right side of column. if ( x != y && top [ x ][ i ] != INT_MAX && bottom [ y ][ i ] != INT_MIN && right [ z ] != INT_MIN ) { ans = max ( ans (( double ) 1 / ( double ) 2 ) * ( bottom [ y ][ i ] - top [ x ][ i ]) * ( right [ z ] - i )); } } return ans ; } // Precompute the vertices of top bottom left // and right and then computing the maximum area. double maxarea ( char mat [ R ][ C ] int r int c ) { int left [ 3 ] right [ 3 ]; int top [ 3 ][ C ] bottom [ 3 ][ C ]; memset ( left INT_MAX sizeof left ); memset ( right INT_MIN sizeof right ); memset ( top INT_MAX sizeof top ); memset ( bottom INT_MIN sizeof bottom ); // finding the r b g cells for the left // and right vertices. for ( int i = 0 ; i < r ; i ++ ) { for ( int j = 0 ; j < c ; j ++ ) { left [ mapcolor ( mat [ i ][ j ])] = min ( left [ mapcolor ( mat [ i ][ j ])] j ); right [ mapcolor ( mat [ i ][ j ])] = max ( left [ mapcolor ( mat [ i ][ j ])] j ); } } // finding set of {r g b} of top and // bottom for each column. for ( int j = 0 ; j < c ; j ++ ) { for ( int i = 0 ; i < r ; i ++ ) { top [ mapcolor ( mat [ i ][ j ])][ j ] = min ( top [ mapcolor ( mat [ i ][ j ])][ j ] i ); bottom [ mapcolor ( mat [ i ][ j ])][ j ] = max ( bottom [ mapcolor ( mat [ i ][ j ])][ j ] i ); } } return findarea ( mat R C top bottom left right ); } // Driven Program int main () { char mat [ R ][ C ] = { 'r' 'r' 'r' 'r' 'r' 'r' 'r' 'r' 'r' 'g' 'r' 'r' 'r' 'r' 'r' 'b' 'b' 'b' 'b' 'b' }; cout < < maxarea ( mat R C ) < < endl ; return 0 ; } Python3import java.util.Arrays ; public class Main { static int R = 4 ; static int C = 5 ; static char [][] mat = { { 'r' 'r' 'r' 'r' 'r' } { 'r' 'r' 'r' 'r' 'g' } { 'r' 'r' 'r' 'r' 'r' } { 'b' 'b' 'b' 'b' 'b' } }; public static void main ( String [] args ) { System . out . println ( maxArea ( mat R C )); } // Returns the color value so that their corresponding index can be accessed. static int mapColor ( char c ) { if ( c == 'r' ) return 0 ; else if ( c == 'g' ) return 1 ; else if ( c == 'b' ) return 2 ; else return - 1 ; } // Returns the maximum area of triangle from all the possible triangles static double findArea ( char [][] mat int r int c int [][] top int [][] bottom int [] left int [] right ) { double ans = 10 ; // For each column for ( int i = 0 ; i < c ; i ++ ) { // For each top vertex for ( int x = 0 ; x < 3 ; x ++ ) { // For each bottom vertex for ( int y = 0 ; y < 3 ; y ++ ) { // Finding the third color of vertex either on right or left. int z = 3 - x - y ; // Finding area of triangle on left side of column. if ( x != y && top [ x ][ i ] != Integer . MAX_VALUE && bottom [ y ][ i ] != Integer . MIN_VALUE && left [ z ] != Integer . MAX_VALUE ) { ans = Math . max ( ans 0.5 * ( bottom [ y ][ i ] - top [ x ][ i ] ) * ( i - left [ z ] )); } // Finding area of triangle on right side of column. if ( x != y && top [ x ][ i ] != Integer . MAX_VALUE && bottom [ y ][ i ] != Integer . MIN_VALUE && right [ z ] != Integer . MIN_VALUE ) { ans = Math . max ( ans 0.5 * ( bottom [ y ][ i ] - top [ x ][ i ] ) * ( right [ z ] - i )); } } } } return ans ; } // Precompute the vertices of top bottom left and right and then computing the maximum area. static double maxArea ( char [][] mat int r int c ) { int [] left = new int [ 3 ] ; Arrays . fill ( left Integer . MAX_VALUE ); int [] right = new int [ 3 ] ; Arrays . fill ( right Integer . MIN_VALUE ); int [][] top = new int [ 3 ][ c ] ; for ( int [] row : top ) Arrays . fill ( row Integer . MAX_VALUE ); int [][] bottom = new int [ 3 ][ c ] ; for ( int [] row : bottom ) Arrays . fill ( row Integer . MIN_VALUE ); // Finding the r b g cells for the left and right vertices. for ( int i = 0 ; i < r ; i ++ ) { for ( int j = 0 ; j < c ; j ++ ) { int color = mapColor ( mat [ i ][ j ] ); left [ color ] = Math . min ( left [ color ] j ); right [ color ] = Math . max ( right [ color ] j ); } } // Finding set of {r g b} of top and bottom for each column. for ( int j = 0 ; j < c ; j ++ ) { for ( int i = 0 ; i < r ; i ++ ) { int color = mapColor ( mat [ i ][ j ] ); top [ color ][ j ] = Math . min ( top [ color ][ j ] i ); bottom [ color ][ j ] = Math . max ( bottom [ color ][ j ] i ); } } return findArea ( mat r c top bottom left right ); } }C## Python3 program to find the maximum # area of triangle having different # vertex color in a matrix. # Return the color value so that their # corresponding index can be access. def mapcolor ( c ): if c == 'r' : return 0 elif c == 'g' : return 1 elif c == 'b' : return 2 # Returns the maximum area of triangle # from all the possible triangles def findarea ( mat r c top bottom left right ): ans = 1 # for each column for i in range ( 0 c ): # for each top vertex for x in range ( 0 3 ): # for each bottom vertex for y in range ( 0 3 ): # finding the third color of # vertex either on right or left. z = 3 - x - y # finding area of triangle on # left side of column. if ( x != y and top [ x ][ i ] != INT_MAX and bottom [ y ][ i ] != INT_MIN and left [ z ] != INT_MAX ): ans = max ( ans 0.5 * ( bottom [ y ][ i ] - top [ x ][ i ]) * ( i - left [ z ])) # finding area of triangle on right side of column. if ( x != y and top [ x ][ i ] != INT_MAX and bottom [ y ][ i ] != INT_MIN and right [ z ] != INT_MIN ): ans = max ( ans 0.5 * ( bottom [ y ][ i ] - top [ x ][ i ]) * ( right [ z ] - i )) return ans # Precompute the vertices of top bottom left # and right and then computing the maximum area. def maxarea ( mat r c ): left = [ - 1 ] * 3 right = [ 0 ] * 3 top = [[ - 1 for i in range ( C )] for j in range ( 3 )] bottom = [[ 0 for i in range ( C )] for j in range ( 3 )] # finding the r b g cells for # the left and right vertices. for i in range ( 0 r ): for j in range ( 0 c ): left [ mapcolor ( mat [ i ][ j ])] = min ( left [ mapcolor ( mat [ i ][ j ])] j ) right [ mapcolor ( mat [ i ][ j ])] = max ( left [ mapcolor ( mat [ i ][ j ])] j ) # finding set of r g b of top # and bottom for each column. for j in range ( 0 c ): for i in range ( 0 r ): top [ mapcolor ( mat [ i ][ j ])][ j ] = min ( top [ mapcolor ( mat [ i ][ j ])][ j ] i ) bottom [ mapcolor ( mat [ i ][ j ])][ j ] = max ( bottom [ mapcolor ( mat [ i ][ j ])][ j ] i ) return int ( findarea ( mat R C top bottom left right )) # Driver Code if __name__ == '__main__' : R C = 4 5 mat = [[ 'r' 'r' 'r' 'r' 'r' ] [ 'r' 'r' 'r' 'r' 'g' ] [ 'r' 'r' 'r' 'r' 'r' ] [ 'b' 'b' 'b' 'b' 'b' ]] INT_MAX INT_MIN = float ( 'inf' ) float ( '-inf' ) print ( maxarea ( mat R C )) # This code is contributed by Rituraj JainJavaScript// C# program to find maximum area of triangle // having different vertex color in a matrix. using System ; class MainClass { const int R = 4 ; const int C = 5 ; // return the color value so that their corresponding // index can be access. static int mapcolor ( char c ) { if ( c == 'r' ) { return 0 ; } else if ( c == 'g' ) { return 1 ; } else if ( c == 'b' ) { return 2 ; } else { return - 1 ; } } // Returns the maximum area of triangle from all // the possible triangles static double findarea ( char [ ] mat int r int c int [ ] top int [ ] bottom int [] left int [] right ) { double ans = . 0 ; // for each column for ( int i = 0 ; i < c ; i ++ ) { // for each top vertex for ( int x = 0 ; x < 3 ; x ++ ) { // for each bottom vertex for ( int y = 0 ; y < 3 ; y ++ ) { // finding the third color of // vertex either on right or left. int z = 3 - x - y ; // finding area of triangle on left side // of column. if ( x != y && top [ x i ] != int . MaxValue && bottom [ y i ] != int . MinValue && left [ z ] != int . MaxValue ) { ans = Math . Max ( ans ( 1.0 / 2.0 ) * ( bottom [ y i ] - top [ x i ]) * ( i - left [ z ])); } // finding area of triangle on right // side of column. if ( x != y && top [ x i ] != int . MaxValue && bottom [ y i ] != int . MinValue && right [ z ] != int . MinValue ) { ans = Math . Max ( ans ( 1.0 / 2.0 ) * ( bottom [ y i ] - top [ x i ]) * ( right [ z ] - i ) + 4 ); } } } } return ans ; } // Precompute the vertices of top bottom left // and right and then computing the maximum area. static double maxarea ( char [ ] mat int r int c ) { int [] left = { int . MaxValue int . MaxValue int . MaxValue }; int [] right = { int . MinValue int . MinValue int . MinValue }; int [ ] top = new int [ 3 C ]; int [ ] bottom = new int [ 3 C ]; // finding the r b g cells for the left // and right vertices. for ( int i = 0 ; i < r ; i ++ ) { for ( int j = 0 ; j < c ; j ++ ) { int color = mapcolor ( mat [ i j ]); if ( color != - 1 ) { left [ color ] = Math . Min ( left [ color ] j ); right [ color ] = Math . Max ( right [ color ] j ); } } } // finding set of {r g b} of top and // bottom for each column. for ( int j = 0 ; j < c ; j ++ ) { for ( int i = 0 ; i < r ; i ++ ) { int color = mapcolor ( mat [ i j ]); if ( color != - 1 ) { top [ color j ] = Math . Min ( top [ color j ] i ); bottom [ color j ] = Math . Max ( bottom [ color j ] i ); } } } return findarea ( mat R C top bottom left right ); } // Driven Program public static void Main ( string [] args ) { char [ ] mat = new char [ ] { { 'r' 'r' 'r' 'r' 'r' } { 'r' 'r' 'r' 'r' 'g' } { 'r' 'r' 'r' 'r' 'r' } { 'b' 'b' 'b' 'b' 'b' } }; Console . WriteLine ( maxarea ( mat R C )); } }Produksjon:
10Lag quiz
Tidskompleksitet: O(R * C)
Hjelpeplass: O(R + C)
Kilde: https://stackoverflow.com/questions/40078660/maximum-area-of-triangle-having-all-vertices-of-different-color
