凸多邊形最優三角剖分
阿新 • • 發佈:2022-05-16
//3d5 凸多邊形最優三角剖分 #include <iostream> using namespace std; const int N = 7;//凸多邊形邊數+1 int weight[][N] = {{0,2,3,1,5,6},{2,0,3,4,8,6},{3,3,0,10,13,5},{1,4,10,0,12,5},{5,8,13,12,0,3},{6,6,7,5,3,0}};//凸多邊形的權 int MinWeightTriangulation(int n,int **t,int **s); void Traceback(int i,int j,int **s);//構造最優解 int Weight(int a,int b,int c);//權函式 int main() { int **s = new int *[N]; int **t = new int *[N]; for(int i=0;i<N;i++) { s[i] = new int[N]; t[i] = new int[N]; } cout<<"此多邊形的最優三角剖分值為:"<<MinWeightTriangulation(N-1,t,s)<<endl; cout<<"最優三角剖分結構為:"<<endl; Traceback(1,5,s); //s[i][j]記錄了Vi-1和Vj構成三角形的第3個頂點的位置 return 0; } int MinWeightTriangulation(int n,int **t,int **s) { for(int i=1; i<=n; i++) { t[i][i] = 0; } for(int r=2; r<=n; r++) //r為當前計算的鏈長(子問題規模) { for(int i=1; i<=n-r+1; i++)//n-r+1為最後一個r鏈的前邊界 { int j = i+r-1;//計算前邊界為r,鏈長為r的鏈的後邊界 t[i][j] = t[i+1][j] + Weight(i-1,i,j);//將鏈ij劃分為A(i) * ( A[i+1:j] )這裡實際上就是k=i s[i][j] = i; for(int k=i+1; k<j; k++) { //將鏈ij劃分為( A[i:k] )* (A[k+1:j]) int u = t[i][k] + t[k+1][j] + Weight(i-1,k,j); if(u<t[i][j]) { t[i][j] = u; s[i][j] = k; } } } } return t[1][N-2]; } void Traceback(int i,int j,int **s) { if(i==j) return; Traceback(i,s[i][j],s); Traceback(s[i][j]+1,j,s); cout<<"三角剖分頂點:V"<<i-1<<",V"<<j<<",V"<<s[i][j]<<endl; } int Weight(int a,int b,int c) { return weight[a][b] + weight[b][c] + weight[a][c]; }