A星算法解决TSP问题

TSP问题相信大部分人已经很熟悉了，就直接展示解法了

#include<iostream>#include<cmath>#include <vector>using namespace std;double dis(int a[], int b[]){    double x1 = a[0];    double y1 = a[1];    double x2 = b[0];    double y2 = b[1];    double x = pow(x1 - x2, 2);    double y = pow(y1 - y2, 2);    return sqrt(x + y);}//double f(int start, int next){////}int main(){    int city[10][2];    for (int i = 0; i < 10; i++)    for (int j = 0; j < 2; j++)        cin >> city[i][j];    int start;    cin >> start;//start-1 = 起始    double discity[10][10];//距离矩阵    for (int i = 0; i < 10; i++)    for (int j = 0; j < 10; j++)    {        if (i == j) discity[i][j] = 999;        else discity[i][j] = dis(city[i], city[j]);    }    int step[20];//用于存储已访问过的城市    int arrived[10] = { 0 };    step[0] = start - 1;    double sum = 0;    int cur = step[0];    arrived[step[0]] = 1;    int next;    int count = 1;    while (count<10) {        cur = step[count - 1];        int min = 9999;        for (int i = 0; i < 10; i++){            if (arrived[i] == 0)            {                if (min > discity[cur][i] + discity[start][i])                {                    min = discity[cur][i] + discity[start][i];                    next = i;                }            }        }        arrived[next] = 1;        sum += discity[cur][next];        //cout << sum << endl;        //cout << discity[cur][next] << endl;        step[count] = next;        count++;        //cout << sum<<endl;    }    sum += discity[cur][next];    for (int i = 0; i < 10;i++) {    cout << step[i] +1<<"->";    }    cout << start << endl;    cout << sum << endl;    //cout << "No output." << endl;    system("pause");    return 0;    //1 2 3 7 4 9 5 1 2 8 4 6 3 8 9 1 8 8 4 4    //5}