TSP-旅行商问题

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <climits>
#include <vector>

#define MAX_N 10

int d[MAX_N][MAX_N];

int dp[1<<MAX_N][MAX_N];

int n;

using namespace std;

int dfs_tsp(int S, int v) {
    if (dp[S][v] >= 0) {
        return dp[S][v];
    }
    if ((1<<n)-1 == S && v == 0) {
        return dp[S][v] = 0;
    }
    
    int res = INT_MAX/2;
    for (int u=0; u<n; u++) {
        if (!((S>>u) & 1)) {
            res = min(res, dfs_tsp(S|(1<<u), u) + d[v][u]);
        }
    }
    return dp[S][v] = res;
}



void dfs_path(int S, int v, vector<int>& path) {
    if ((1<<n)-1 == S && v == 0) {
        return;
    }
    
    int res = INT_MAX / 2;
    int idx = -1;
    for (int u=0; u<n; u++) {
        int cdst = dp[S|(1<<u)][u] + d[v][u];
        if (!((S>>u) & 1) && cdst < res) {
            idx = u;
            res = cdst;
        }
    }
    path.push_back(idx);
    dfs_path(S|(1<<idx), idx, path);
}

int main() {
    n = 5;
    memset(dp, -1, sizeof(dp));

    for (int i=0; i<n; i++) {
        for (int j=0; j<n; j++) {
            d[i][j] = INT_MAX/2;
        }
    }
    
    d[0][1] = 3;
    d[0][3] = 4;
    d[4][0] = 7;
    d[4][1] = 6;
    
    d[3][4] = 3;
    d[2][0] = 4;
    d[2][3] = 5;
    d[1][2] = 5;
    
    
    
    int dst = dfs_tsp(0, 0);
    cout<<dst<<endl;

    vector<int> path(1,0);

    dfs_path(0, 0, path);
    
    for (int e:path) {
        cout<<e<<" ";
    }

    system("pause");
    return 0;
}

挑战程序设计P193. 其实还可以反向着来，不过实现起来比这个稍烦一些。