Maksimālais trīsstūra laukums ar dažādām virsotņu krāsām
Dota matrica no N rindas un M kolonnas sastāv no trim vērtībām {r g b}. Uzdevums ir atrast laukumu lielākajam trīsstūrim, kura viena mala ir paralēla y asij, t.i., vertikāla, un visu trīs virsotņu krāsa ir atšķirīga.
Piemēri:
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.
Mēs zinām trīsstūra laukumu = 1/2 * bāze * augstums, tāpēc mums ir jāpalielina trīsstūra pamatne un augstums. Tā kā viena mala ir paralēla y asij, mēs to varam uzskatīt par trijstūra pamatu.
C++
Lai palielinātu bāzi, katrai kolonnai varam atrast pirmo un pēdējo {r g b} gadījumu. Tātad katrai kolonnai ir divas 3 vērtību kopas. Jebkuras kolonnas bāzei viena virsotne ir no pirmās kopas un otrā virsotne no otrās kopas tā, ka tām ir dažādas vērtības.
Lai palielinātu jebkuras kolonnas kā pamatnes augstumu, trešā virsotne ir jāizvēlas tā, lai virsotne būtu vistālāk no kolonnas kolonnas kreisajā vai labajā pusē, kuras vērtība atšķiras no pārējām divām virsotnēm.
Tagad katrai kolonnai atrodiet maksimālo trīsstūra laukumu.
Tālāk ir sniegta informācija par šīs pieejas īstenošanu.
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 )); } }Izvade:
10Izveidojiet viktorīnu
Laika sarežģītība: O(R*C)
Palīgtelpa: O(R+C)
Avots: https://stackoverflow.com/questions/40078660/maximum-area-of-triangle-having-all-vertices-of-different-color
