给定一个N行M列的矩阵，它由三个值{r，g，b}组成。任务是找到最大一侧与y轴平行(即垂直)的最大三角形的面积，并且所有三个顶点的颜色都不同。
例子：
输入： N = 4，M = 5 mat [] [] = {r，r，r，r，r，r，r，r，g，r，r，r，r，r，b，b， b，b，b，}输出： 10三角形的最大面积为10。三角形坐标为(0,0)包含r，(1,4)包含g，(3,0)包含b。 
我们知道三角形的面积= 1/2 *基数*高，因此我们需要最大化三角形的基数和高度。由于一侧平行于y轴，因此我们可以将该侧视为三角形的底。
为了最大化基数，我们可以找到每一列的{r，g，b}的第一个和最后一个出现。因此，每列有两组3个值。对于任何列中的基，一个顶点来自第一组，第二顶点来自第二组，以使它们具有不同的值。
为了使高度最大化，对于任何以列为基础的列，必须选择第三个顶点，以使该顶点离列最远，并且在列的左侧或右侧具有与其他两个顶点不同的值。
现在，对于每一列，找到三角形的最大面积。
以下是此方法的实现：
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);
}
}
// finsing 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;
} 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)
# finsing 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输出：
10
时间复杂度： O(M * N)。
资料来源： http：//stackoverflow.com/questions/40078660/maximum-area-of-triangle-having-all-vertices-of-different-color