Dada una matriz de norte filas y METRO Las columnas constan de tres valores {r g b}. La tarea es encontrar el área del triángulo más grande que tiene un lado paralelo al eje y, es decir, vertical, y el color de los tres vértices es diferente.
Ejemplos:
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.
Sabemos que el área de un triángulo = 1/2 * base * altura, por lo que debemos maximizar la base y la altura del triángulo. Como un lado es paralelo al eje y, podemos considerar ese lado como la base del triángulo.
Para maximizar la base podemos encontrar la primera y la última aparición de {r g b} para cada columna. Entonces tenemos dos conjuntos de 3 valores para cada columna. Para la base de cualquier columna, un vértice es del primer conjunto y el segundo vértice del segundo conjunto, de modo que tienen valores diferentes.
Para maximizar la altura de cualquier columna como base, el tercer vértice debe elegirse de modo que el vértice esté más alejado de la columna en el lado izquierdo o derecho de la columna y tenga un valor diferente de los otros dos vértices.
Ahora, para cada columna, encuentra el área máxima del triángulo.
A continuación se muestra la implementación de este enfoque:
C++
// C++ program to find maximum area of triangle // having different vertex color in a matrix. #include
import 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); } }
# 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 Jain
// 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)); } }
// Javascript program to find maximum area of triangle // having different vertex color in a matrix. // return the color value so that their corresponding // index can be accessed. function mapcolor(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 function findarea(mat r c top bottom left right) { let ans = 10; // for each column for (let i = 0; i < c; i++) { // for each top vertex for (let x = 0; x < 3; x++) { // for each bottom vertex for (let y = 0; y < 3; y++) { // finding the third color of // vertex either on right or left. let z = 3 - x - y; // finding area of triangle on left side of column. if (x != y && top[x][i] != Number.MAX_SAFE_INTEGER && bottom[y][i] != Number.MIN_SAFE_INTEGER && left[z] != Number.MAX_SAFE_INTEGER) { ans = Math.max(ans (1/2) * (bottom[y][i] - top[x][i]) * (i - left[z])); } // finding area of triangle on right side of column. if (x != y && top[x][i] != Number.MAX_SAFE_INTEGER && bottom[y][i] != Number.MIN_SAFE_INTEGER && right[z] != Number.MIN_SAFE_INTEGER) { ans = Math.max(ans (1/2) * (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. function maxarea(mat r c) { let left = [Number.MAX_SAFE_INTEGER Number.MAX_SAFE_INTEGER Number.MAX_SAFE_INTEGER]; let right = [Number.MIN_SAFE_INTEGER Number.MIN_SAFE_INTEGER Number.MIN_SAFE_INTEGER]; let top = Array.from({length: 3} () => Array(c).fill(Number.MAX_SAFE_INTEGER)); let bottom = Array.from({length: 3} () => Array(c).fill(Number.MIN_SAFE_INTEGER)); // finding the r b g cells for the left // and right vertices. for (let i = 0; i < r; i++) { for (let j = 0; j < c; j++) { let 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 (let j = 0; j < c; j++) { for (let i = 0; i < r; i++) { let 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); } // Driven Program const R = 4; const C = 5; const mat = [ ['r' 'r' 'r' 'r' 'r'] ['r' 'r' 'r' 'r' 'g'] ['r' 'r' 'r' 'r' 'r'] ['b' 'b' 'b' 'b' 'b'] ]; console.log(maxarea(mat R C)); // akashish__
Producción:
10
Complejidad del tiempo: O(R*C)
Espacio Auxiliar: O(R+C)
Fuente: https://stackoverflow.com/questions/40078660/área-máxima-de-triángulo-que-tiene-todos-los-vértices-de-diferente-color