最小成本多边形三角剖分

凸多边形的三角剖分是通过在不相邻的顶点（角）之间绘制对角线来形成的，这样对角线就不会相交。问题是找到成本最低的三角剖分成本。三角剖分的成本是其组成三角形的权重之和。每个三角形的重量是它的周长（所有边长之和）
请参阅取自此来源的以下示例。

同一个凸五边形的两个三角剖分。左侧三角剖分的成本为 8 + 2√2 + 2√5（约 15.30），右侧三角剖分的成本为 4 + 2√2 + 4√5（约 15.77）。

这个问题具有递归子结构。这个想法是将多边形分成三部分：单个三角形，左侧的子多边形和右侧的子多边形。我们像这样尝试所有可能的划分，并找到最小化三角形成本加上两个子多边形三角剖分成本的划分。

Let Minimum Cost of triangulation of vertices from i to j be minCost(i, j)
If j < i + 2 Then
  minCost(i, j) = 0
Else
  minCost(i, j) = Min { minCost(i, k) + minCost(k, j) + cost(i, k, j) }
                  Here k varies from 'i+1' to 'j-1'

Cost of a triangle formed by edges (i, j), (j, k) and (k, i) is 
  cost(i, j, k)  = dist(i, j) + dist(j, k) + dist(k, i)

以下是上述朴素递归公式的实现。

C++

// Recursive implementation for minimum cost convex polygon triangulation
#include 
#include 
#define MAX 1000000.0
using namespace std;
 
// Structure of a point in 2D plane
struct Point
{
    int x, y;
};
 
// Utility function to find minimum of two double values
double min(double x, double y)
{
    return (x <= y)? x : y;
}
 
// A utility function to find distance between two points in a plane
double dist(Point p1, Point p2)
{
    return sqrt((p1.x - p2.x)*(p1.x - p2.x) +
                (p1.y - p2.y)*(p1.y - p2.y));
}
 
// A utility function to find cost of a triangle. The cost is considered
// as perimeter (sum of lengths of all edges) of the triangle
double cost(Point points[], int i, int j, int k)
{
    Point p1 = points[i], p2 = points[j], p3 = points[k];
    return dist(p1, p2) + dist(p2, p3) + dist(p3, p1);
}
 
// A recursive function to find minimum cost of polygon triangulation
// The polygon is represented by points[i..j].
double mTC(Point points[], int i, int j)
{
   // There must be at least three points between i and j
   // (including i and j)
   if (j < i+2)
      return 0;
 
   // Initialize result as infinite
   double res = MAX;
 
   // Find minimum triangulation by considering all
   for (int k=i+1; k

---

Java

// Class to store a point in the Euclidean plane
class Point
{
  int x, y;
  public Point(int x, int y)
  {
    this.x = x;
    this.y = y;
  }
 
  // Utility function to return the distance between two
  // vertices in a 2-dimensional plane
  public double dist(Point p)
  {
 
    // The distance between vertices `(x1, y1)` & `(x2,
    // y2)` is `√((x2 − x1) ^ 2 + (y2 − y1) ^ 2)`
    return Math.sqrt((this.x - p.x) * (this.x - p.x)
                     + (this.y - p.y) * (this.y - p.y));
  }
}
 
class GFG
{
 
  // Function to calculate the weight of optimal
  // triangulation of a convex polygon represented by a
  // given set of vertices `vertices[i..j]`
  public static double MWT(Point[] vertices, int i, int j)
  {
 
    // If the polygon has less than 3 vertices,
    // triangulation is not possible
    if (j < i + 2)
    {
      return 0;
    }
 
    // keep track of the total weight of the minimum
    // weight triangulation of `MWT(i,j)`
    double cost = Double.MAX_VALUE;
 
    // consider all possible triangles `ikj` within the
    // polygon
    for (int k = i + 1; k <= j - 1; k++)
    {
 
      // The weight of a triangulation is the length
      // of perimeter of the triangle
      double weight = vertices[i].dist(vertices[j])
        + vertices[j].dist(vertices[k])
        + vertices[k].dist(vertices[i]);
 
      // choose the vertex `k` that leads to the
      // minimum total weight
      cost = Double.min(cost,
                        weight + MWT(vertices, i, k)
                        + MWT(vertices, k, j));
    }
    return cost;
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    // vertices are given in clockwise order
    Point[] vertices
      = { new Point(0, 0), new Point(2, 0),
         new Point(2, 1), new Point(1, 2),
         new Point(0, 1) };
 
    System.out.println(MWT(vertices,
                           0, vertices.length - 1));
  }
}
 
// This code is contributed by Priiyadarshini Kumari

Python3

# Recursive implementation for minimum
# cost convex polygon triangulation
from math import sqrt
MAX = 1000000.0
 
# A utility function to find distance
# between two points in a plane
def dist(p1, p2):
    return sqrt((p1[0] - p2[0])*(p1[0] - p2[0]) + \
                (p1[1] - p2[1])*(p1[1] - p2[1]))
 
# A utility function to find cost of
# a triangle. The cost is considered
# as perimeter (sum of lengths of all edges)
# of the triangle
def cost(points, i, j, k):
    p1 = points[i]
    p2 = points[j]
    p3 = points[k]
    return dist(p1, p2) + dist(p2, p3) + dist(p3, p1)
 
 
# A recursive function to find minimum
# cost of polygon triangulation
# The polygon is represented by points[i..j].
def mTC(points, i, j):
     
    # There must be at least three points between i and j
    # (including i and j)
    if (j < i + 2):
        return 0
         
    # Initialize result as infinite
    res = MAX
     
    # Find minimum triangulation by considering all
    for k in range(i + 1, j):
        res = min(res, (mTC(points, i, k) + \
                        mTC(points, k, j) + \
                        cost(points, i, k, j)))
     
    return round(res, 4)
 
 
# Driver code
points = [[0, 0], [1, 0], [2, 1], [1, 2], [0, 2]]
n = len(points)
print(mTC(points, 0, n-1))
 
# This code is contributed by SHUBHAMSINGH10

输出：

15.3006

上述问题类似于矩阵链乘法。下面是 mTC(points[], 0, 4) 的递归树。

从上面的递归树中可以很容易看出，这个问题有很多重叠的子问题。由于该问题具有两个性质：最优子结构和重叠子问题，因此可以使用动态规划有效地解决它。
以下是动态编程解决方案的 C++ 实现。

C

// A Dynamic Programming based program to find minimum cost of convex
// polygon triangulation
#include 
#include 
#define MAX 1000000.0
using namespace std;
 
// Structure of a point in 2D plane
struct Point
{
    int x, y;
};
 
// Utility function to find minimum of two double values
double min(double x, double y)
{
    return (x <= y)? x : y;
}
 
// A utility function to find distance between two points in a plane
double dist(Point p1, Point p2)
{
    return sqrt((p1.x - p2.x)*(p1.x - p2.x) +
                (p1.y - p2.y)*(p1.y - p2.y));
}
 
// A utility function to find cost of a triangle. The cost is considered
// as perimeter (sum of lengths of all edges) of the triangle
double cost(Point points[], int i, int j, int k)
{
    Point p1 = points[i], p2 = points[j], p3 = points[k];
    return dist(p1, p2) + dist(p2, p3) + dist(p3, p1);
}
 
// A Dynamic programming based function to find minimum cost for convex
// polygon triangulation.
double mTCDP(Point points[], int n)
{
   // There must be at least 3 points to form a triangle
   if (n < 3)
      return 0;
 
   // table to store results of subproblems.  table[i][j] stores cost of
   // triangulation of points from i to j.  The entry table[0][n-1] stores
   // the final result.
   double table[n][n];
 
   // Fill table using above recursive formula. Note that the table
   // is filled in diagonal fashion i.e., from diagonal elements to
   // table[0][n-1] which is the result.
   for (int gap = 0; gap < n; gap++)
   {
      for (int i = 0, j = gap; j < n; i++, j++)
      {
          if (j < i+2)
             table[i][j] = 0.0;
          else
          {
              table[i][j] = MAX;
              for (int k = i+1; k < j; k++)
              {
                double val = table[i][k] + table[k][j] + cost(points,i,j,k);
                if (table[i][j] > val)
                     table[i][j] = val;
              }
          }
      }
   }
   return  table[0][n-1];
}
 
// Driver program to test above functions
int main()
{
    Point points[] = {{0, 0}, {1, 0}, {2, 1}, {1, 2}, {0, 2}};
    int n = sizeof(points)/sizeof(points[0]);
    cout << mTCDP(points, n);
    return 0;
}

输出：

15.3006

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。