算法设计与分析——回溯法——旅行售货员问题

#include<iostream>
#include<bits/stdc++.h>
using namespace std;
const int noEdge=65535;
class Traveling
{
	public:
		void BackTrack(int i);
		
		int n;		//图G的顶点数 
		int *x;		//当前的解 
		int *bestx; // 当前的最优解 
		int **a;    // 图G的临界矩阵 
		int cc;     // 当前费用 
		int bestc;  //当前最优花费的值 
	//  int noEdge  无边标记 
 };
 
 void Traveling::BackTrack(int i)
 {
 	if(i == n)
 	{
 		if(a[x[n-1]][x[n]] !=noEdge && a[x[n]][1] !=noEdge 
		   &&(cc+a[x[n-1]][x[n]]+a[x[n]][1]<bestc || bestc ==noEdge))
		   {
		   	for(int j=1;j<=n;j++)
		   	{
		   		bestx[j]=x[j];
			}
			bestc=cc+a[x[n-1]][x[n]]+a[x[n]][1];
		   }
	 }
	 else
	 {
	 	for(int j=i;j<=n;j++)
	 	{
	 		if(a[x[i-1]][x[j]] != noEdge && (cc+a[x[i-1]][x[j]]<bestc||bestc == noEdge))
	 		{
	 			swap(x[i],x[j]);
	 			cc+=a[x[i-1]][x[i]];
	 			BackTrack(i+1);
	 			cc-=a[x[i-1]][x[i]];
				swap(x[i],x[j]);
			 }
		 }
	 }
  } 
  
  int TSP(int **a,int v[],int n)
  {
  	Traveling Y;
  	Y.x = new int [n+1];
  	for(int i=1;i<=n;i++)
  	{
  		Y.x[i]=i;
	  }
	  Y.a= a;
	  Y.n= n;
	  Y.bestc = noEdge;
	  Y.bestx = v;
	  Y.cc = 0;
	  
	  Y.BackTrack(2);
	  cout<<"旅行售货员问题的最优路径：";
	  for(int i=1;i<=n;i++)
	{
		cout<<Y.bestx[i]<<"  ";
		 
	}
	cout<<1<<endl;
	  delete []Y.x;
	  return Y.bestc; 
	  
  }
int main()
{
	int n;
	
	cout<<"输入城市的个数";
	cin>>n;
	int **vector = new int *[n+1];
	for(int i=1;i<=n;i++)
	{
		vector[i] = new int [n+1];
	}
	for(int i=1;i<=n;i++)
	 {
	 	for(int j=1;j<=n;j++)
	 	{
	 		vector[i][j]= noEdge;
		 }
	
	 }
	int x;
	cout<<"输入城市间存在的通路的个数";
	cin>>x;
	cout<<"输入数据的形式为：(1 2 3)"<<endl;
	int index_i,index_j,value;
	 for(int i=0;i<x;i++)
	 {
	 	cin>>index_i>>index_j>>value;
	 	vector[index_i][index_j]=value;
	 	vector[index_j][index_i]=value;
	 }
	 cout<<"输出临界矩阵"<<endl; 
	 for(int i=1;i<=n;i++)
	 {
	 	for(int j=1;j<=n;j++)
	 	{
	 		cout<<setw(5)<<vector[i][j]<<"  ";
		 }
		 cout<<endl;
	 }
	 int *v = new int [n+1];
	 
	 
	 
	cout<<"旅行售货员问题的最优值为:"<<TSP(vector,v,n)<<endl;
	
	
 }