Givet en matrix af N rækker og M kolonner består af tre værdier {r g b}. Opgaven er at finde arealet af den største trekant, der har en side parallel med y-aksen, dvs. lodret, og farven på alle tre hjørner er forskellige.
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 kender arealet af en trekant = 1/2 * base *højde, så vi skal maksimere trekantens base og højde. Da den ene side er parallel med y-aksen, kan vi betragte den side som trekantens base.
For at maksimere basis kan vi finde den første og sidste forekomst af {r g b} for hver kolonne. Så vi har to sæt med 3 værdier for hver kolonne. For base i enhver kolonne er et toppunkt fra det første sæt og det andet toppunkt fra det andet sæt, således at de har forskellige værdier.
For at maksimere højden for enhver søjle som basis skal det tredje toppunkt vælges således, at toppunktet skal være længst væk fra søjlen på venstre eller højre side af søjlen med en værdi forskellig fra de to andre toppunkter.
Find nu trekantens maksimale areal for hver kolonne.
Nedenfor er implementeringen af denne tilgang:
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__
Produktion:
10
Tidskompleksitet: O(R * C)
Hjælpeplads: O(R + C)
Kilde: https://stackoverflow.com/questions/40078660/maximum-area-of-triangle-having-all-vertices-of-different-color