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Kolmion enimmäispinta-ala, jossa on eri kärkivärejä

Annettu matriisi N rivit ja M sarakkeet koostuu kolmesta arvosta {r g b}. Tehtävänä on löytää pinta-ala suurimmasta kolmiosta, jonka toinen sivu on yhdensuuntainen y-akselin kanssa eli pystysuora ja jonka kaikkien kolmen kärjen väri on erilainen.
Esimerkkejä:  
 

    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.

Kolmion enimmäispinta-ala, jossa on eri kärkivärejä

 

Tiedämme kolmion pinta-alan = 1/2 * kanta *korkeus, joten meidän täytyy maksimoida kolmion kanta ja korkeus. Koska toinen sivu on yhdensuuntainen y-akselin kanssa, voimme pitää sitä sivua kolmion kantana.
Kantakohdan maksimoimiseksi voimme löytää kunkin sarakkeen {r g b}:n ensimmäisen ja viimeisen esiintymän. Meillä on siis kaksi kolmen arvon sarjaa jokaiselle sarakkeelle. Minkä tahansa sarakkeen kannalle yksi kärkipiste on ensimmäisestä joukosta ja toinen kärki toisesta joukosta siten, että niillä on eri arvot.
Minkä tahansa sarakkeen korkeuden maksimoimiseksi pohjana kolmas kärki on valittava siten, että kärjen tulee olla kauimpana sarakkeesta sarakkeen vasemmalla tai oikealla puolella, jonka arvo on erilainen kuin kahdella muulla kärjellä. 
Etsi nyt jokaisesta sarakkeesta kolmion enimmäispinta-ala.
Alla on tämän lähestymistavan toteutus:
 



C++
// C++ program to find maximum area of triangle // having different vertex color in a matrix. #include   using 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; } 
Java
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
# 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#
// 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
// 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__ 

Lähtö:  

10


Aika monimutkaisuus: O(R * C)
Aputila: O(R + C) 
Lähde: https://stackoverflow.com/questions/40078660/maximum-area-of-triangle-having-all-vertices-of-different-color
 

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