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//3d5 凸多边形最优三角剖分
#include "stdafx.h"
#include <iostream> 
using namespace std; 
 
const int N = 7;//凸多边形边数+1
int weight[][N] = {{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}};//凸多边形的权
 
int MinWeightTriangulation(int n,int **t,int **s);
void Traceback(int i,int j,int **s);//构造最优解
int Weight(int a,int b,int c);//权函数
 
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<<"此多边形的最优三角剖分值为："<<MinWeightTriangulation(N-1,t,s)<<endl;  
    cout<<"最优三角剖分结构为："<<endl;  
    Traceback(1,5,s); //s[i][j]记录了Vi-1和Vj构成三角形的第3个顶点的位置
 
	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为当前计算的链长（子问题规模）  
	{
		for(int i=1; i<=n-r+1; i++)//n-r+1为最后一个r链的前边界  
		{
			int j = i+r-1;//计算前边界为r，链长为r的链的后边界  
 
			t[i][j] = t[i+1][j] + Weight(i-1,i,j);//将链ij划分为A(i) * ( A[i+1:j] )这里实际上就是k=i
 
			s[i][j] = i;
 
			for(int k=i+1; k<j; k++)
			{
				//将链ij划分为( 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<<"三角剖分顶点：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];
}