The problem of optimal triangular division of Convex Polygon

See the book "algorithm design and analysis" Wang Xiaodong Dynamic Planning 1. problem description (Note: it is the sum of weights of all triangles, not the sum of weights of edges and chords) 2. analysis 3. encoding implementation :/**
* @ Author: Hu Jiawei
* @ Createtime: 12:31:16
* @ Description: optimal triangular division of a convex polygon.
*/

PackageEx2;

Public ClassTriangulation {

Private IntN; // n Polygon
Private Int[] [] Weight; // edge weight array

PublicTriangulation (IntN ){
This. N = N;
This. Weight =New Int[N] [N];
}

Public Static VoidMain (string [] ARGs ){
Triangulation triangulation =NewTriangulation (6 );
Inittriangulation (Triangulation );
IntN = triangulation. getn (); // Number of edges of a convex polygon
Int[] [] T =New Int[N] [N]; // T [I] [J] indicates the weight of the optimal triangle partitioning of a polygon consisting of a vertex {Vi-1, vi... VJ}
Int[] [] S =New Int[N] [N]; // s [I] [J] represents the position of the third vertex of the triangle together with the Vi-1 and vj
Triangulation. minweighttriangulation2 (triangulation. getn ()-1, t, s );
System. Out. println (T [1] [5]);
}

// Initialize weight array information
Public Static VoidInittriangulation (triangulation Triangulation ){
Int[] [] Weight = {0, 2, 2, 3, 1, 4}, {2, 0, 1, 5, 2, 3}, {2, 1, 0, 2, 1, 4 },
{3, 5, 2, 0, 6, 2}, {1, 2, 1, 6, 0, 1}, {4, 3, 4, 2, 1, 0 }};
Triangulation. setweight (weight );
}

// Obtain the optimal triangle partitioning. n is the total number of edges-1.
Public VoidMinweighttriangulation (IntN,Int[] [] T,Int[] [] S ){
// Initialize all the two-vertex polygon with a weight of 0
For(IntI = 1; I <= N; I ++ ){
T [I] [I] = 0;
}
// Cyclically solving T [I] [J]
For(IntR = 2; r <= N; r ++) {// (J-I) range [2, N]
// When r = 2, the loop is actually assigning edge values to T, that is, the weights of two adjacent vertices, such as t [1] [2], T [2] [3]...
For(IntI = 1; I <= N-R + 1; I ++) {// I range [1, N + 1-r], here I want to ensure I + r <= N
IntJ = I + R-1;
T [I] [J] = T [I + 1] [J] + getweight (I-1, I, j); // here actually K = I
// T [I] [J] = T [I] [I] + T [I + 1] [J] + getweight (I-1, I, j)
S [I] [J] = I;
// I-1, I, j
// Loop K, range is [I + 1, J-1], find the smallest T [I] [J]
For(IntK = I + 1; k <j; k ++) {// K is the intermediate vertex between I and j
IntU = T [I] [k] + T [k + 1] [J] + getweight (I-1, K, J ); // use K as the weight.
If(U <t [I] [J]) {// if the weight is smaller, update T [I] [J] And s [I] [J] at the same time.
T [I] [J] = u;
S [I] [J] = K;
}
}
}
}
}

// My writing method is different in the second loop. There is no difference, but it is easy for me to understand.
Public VoidMinweighttriangulation2 (IntN,Int[] [] T,Int[] [] S ){
// Initialize all the two-vertex polygon with a weight of 0
For(IntI = 1; I <= N; I ++ ){
T [I] [I] = 0;
}
// Cyclically solving T [I] [J]
For(IntR = 1; r <= N; r ++) {// R = (J-I) range [1, N]
// When r = 1, the loop is actually assigning edge values to T, that is, the weights of two adjacent vertices, such as t [1] [2], T [2] [3], t [3] [4]...
For(IntI = 1; I <= N-R; I ++) {// I range [1, n-R], here I want to ensure that J = I + r <= N
IntJ = I + R;
T [I] [J] = T [I + 1] [J] + getweight (I-1, I, j); // here actually K = I
// T [I] [J] = T [I] [I] + T [I + 1] [J] + getweight (I-1, I, j)
S [I] [J] = I; // I-1, I, j
// Loop K, range is [I + 1, J-1], find the smallest T [I] [J]
For(IntK = I + 1; k <j; k ++) {// K is the intermediate vertex between I and j
IntU = T [I] [k] + T [k + 1] [J] + getweight (I-1, K, J ); // use K as the weight.
If(U <t [I] [J]) {// if the weight is smaller, update T [I] [J] And s [I] [J] at the same time.
T [I] [J] = u;
S [I] [J] = K;
}
}
}
}
}

// Calculate the sum of the weights of a triangle
Public IntGetweight (IntI,IntJ,IntK ){
ReturnWeight [I] [J] + weight [J] [k] + weight [I] [k];
}

Public IntGetn (){
ReturnN;
}

Public VoidSetn (IntN ){
This. N = N;
}

Public Int[] [] Getweight (){
ReturnWeight;
}

Public VoidSetweight (Int[] [] Weight ){
This. Weight = weight;
}

}

Data: Result: 24

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