Backtracking to solve the shortest path

Use backtracking to solve the following problem

Question: a salesman to a number of cities to sell commodities, the known distance between cities. He selected from a resident, after each city again, and finally back to the station's route to minimize total distance (or the total travel expenses). The number of cities less than five.

 

 

#include<iostream>
using namespace std;

const int INF = 10000000;
int n, cc = 0, bestc = INF;
int **g;
int *x, *bestx;

void travel(int t) {
    if (t == n) {
        if (g[x[t - 1]][x[t]] != INF && g[x[t]][1] != INF &&
            (cc + g[x[t - 1]][x[t]] + g[x[t]][1] < bestc || bestc == INF)) {
            for (int i = 0; i < n + 1; i++)
                bestx[i] = x[i];
            bestc = cc + g[x[t - 1]][x[t]] + g[x[t]][1];
        }
        return;
    }

    for (int i = t; i < n; i++) {
        if (g[x[t - 1]][x[i]] != INF && (cc + g[x[t - 1]][x[i]] < bestc
            || bestc == INF)) {
            swap(x[i], x[t]);
            cc += g[x[t - 1]][x[t]];
            travel(t + 1);
            cc -= g[x[t - 1]][x[t]];
            swap(x[i], x[t]);
        }
    }
}

void output() {
    cout << "最短路程为：" << " ";
    cout << bestc << endl;
    cout << "最优路径为：" << " ";
    cout << bestx[1];
    for (int i = 2; i < n + 1; I ++)
        COUT << "  " << bestx [I]; 
    COUT << "  " << bestx [ . 1 ] << endl; 
} 

int main () { 

    n = . 4 ; // set the number n of the city. 4 
    G = new new  int * [n-+ . 1 ]; // G is represented by an array of storage path 
    X = new new  int [n-+ . 1 ]; 
    bestx = new new  int [n-+ . 1 ]; 

    for ( int I = 0; i < n + 1; i++) {
        g[i] = new int[n + 1];
        x[i] = i;

        for (int j = 0; j < n + 1; j++)
            g[i][j] = INF;
    }

    g[1][2] = g[2][1] = 30;//1<-->2之间距离
    g[1][3] = g[3][1] = 6;
    g[1][4] = g[4][1] = 4;

    g[2][3] = g[3][2] = 5;
    g[2][4] = g[4][2] = 10;

    g[3][4] = g[4][3] = 20;

    travel(2);
    output();
    system("pause");
    return 0;
}