Največja površina trikotnika z različnimi barvami oglišč
Glede na matriko n vrstice in M stolpec je sestavljen iz treh vrednosti {r g b}. Naloga je najti ploščino največjega trikotnika, ki ima eno stran vzporedno z osjo y, tj. navpično, barve vseh treh oglišč pa so različne.
Primeri:
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.
Vemo, da je površina trikotnika = 1/2 * osnova * višina, zato moramo povečati osnovo in višino trikotnika. Ker je ena stranica vzporedna z osjo y, jo lahko štejemo za osnovo trikotnika.
C++
Za povečanje baze lahko poiščemo prvo in zadnjo pojavitev {r g b} za vsak stolpec. Tako imamo dva niza po 3 vrednosti za vsak stolpec. Za osnovo v katerem koli stolpcu je eno oglišče iz prvega niza in drugo oglišče iz drugega niza, tako da imata različne vrednosti.
Za maksimiranje višine katerega koli stolpca kot osnove mora biti tretje oglišče izbrano tako, da mora biti oglišče najbolj oddaljeno od stolpca na levi ali desni strani stolpca z vrednostjo, ki se razlikuje od drugih dveh oglišč.
Zdaj za vsak stolpec poiščite največjo površino trikotnika.
Spodaj je izvedba tega pristopa:
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 )); } }Izhod:
10Ustvari kviz
Časovna zapletenost: O(R * C)
Pomožni prostor: O(R + C)
Vir: https://stackoverflow.com/questions/40078660/maximum-area-of-triangle-having-all-vertices-of-different-color
