双调旅行售货员问题

#include "iostream"
#include "cmath"
#include "fstream"
#include "algorithm"
using namespace std;

/*
t(i)表示点{p1,p2...pi}的最短双调TSP回路，其中p1,p2..pi按x坐标排序
t(i) = min{t(k) + D(k,i) + d(k-1,i) - d(k-1,k)}  1<k<i
t(1) = 0   t(1) = 2d(1,2)
d(i,j)为pi,pj之间的距离
D(i,j)为pi,pi+1,pi+1...pj之间累加距离

设s(i)=d(1,2) + d(2,3) +...+ d(i-1,i)
则D(k, i) = s(k) - s(i); d(k-1, k) = s(k) - s(k-1)
所以t(i) = min{t(k) + s(i) + s(k-1) - 2s(k) + d(k-1, i)}  1<k<i
*/

const int MAX = 100;
double s[MAX];
double t[MAX];
int N;  //点的数量

struct Point
{
int x;  //x坐标
int y;  //y坐标

bool operator < (const Point &p) const
{
return x < p.x;
}
};

Point point[MAX];

double dist(Point p1, Point p2) //两点之间的距离
{
return sqrt( pow((p1.x-p2.x),2) + pow((p1.y-p2.y),2) );
}

//t(i) = min{t(k) + s(i) + s(k-1) - 2s(k) + d(k-1, i)}  1<k<i
double dyna()
{
sort(point, point+N);  //按x坐标排序
int i, k;
s[1] = 0;
for(i=2; i<=N; i++)
s[i] = s[i-1] + dist(point[i-1], point[i]);

t[1] = 0;
t[2] = 2 * dist(point[1], point[2]);
double temp;
for(i=3; i<=N; i++)
{
t[i] = t[2] + s[i] + s[1] - 2*s[2] + dist(point[1], point[i]);
for(k=3; k<i; k++)
{
temp = t[k] + s[i] + s[k-1] - 2*s[k] + dist(point[k-1], point[i]);
if(temp < t[i])
t[i] = temp;
}
}
return t
;
}

int main()
{
ifstream fin("旅行.txt");
cout << "输入点的数量：";
fin >> N; cout << N;
cout << "\n输入各点的坐标：\n";
int i;
for(i=1; i<=N; i++)
{
fin >> point[i].x >> point[i].y;
cout << point[i].x << " " << point[i].y << endl;
}
cout << "最短双调TSP回路长度为：" << dyna() << endl;
fin.close();
return 0;
}