UG二次开发中利用open API的函数以及蚁群算法实现选取N个点，搜寻着N个点之间的最短路径并且连接起来用C++代码实现

以下是使用open API的函数以及蚁群算法实现选取N个点之间的最短路径并连接起来的C++代码示例：

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <ctime>

using namespace std;

// 定义城市结构体
struct City {
    int id;
    double x;
    double y;
    City(int i, double xx, double yy) : id(i), x(xx), y(yy) {}
};

// 定义蚂蚁结构体
struct Ant {
    vector<int> tour; // 蚂蚁已经走过的路径
    vector<bool> visited; // 标记某个城市是否被访问过
    double length; // 蚂蚁走过的路径长度
    Ant(int n) : visited(n, false), length(0.0) {}
};

// 定义全局变量
const double ALPHA = 1.0; // 信息素重要程度因子
const double BETA = 5.0; // 启发式因子
const double RHO = 0.5; // 信息素挥发因子
const int Q = 100; // 信息素增量常数
const int MAX_ITERATIONS = 100; // 最大迭代次数
const double MAX_DISTANCE = 100000.0; // 两个城市之间的最大距离

// 计算两个城市之间的距离
double distance(City& a, City& b) {
    double dx = a.x - b.x;
    double dy = a.y - b.y;
    return sqrt(dx * dx + dy * dy);
}

// 计算一条路径的总长度
double pathLength(vector<int>& tour, vector<City>& cities) {
    double result = 0.0;
    int n = tour.size();
    for (int i = 0; i < n; ++i) {
        int j = (i == n - 1) ? 0 : i + 1;
        result += distance(cities[tour[i]], cities[tour[j]]);
    }
    return result;
}

// 计算蚂蚁从某个城市到其他城市的概率
void probabilities(int currentCity, vector<int>& candidateCities, vector<City>& cities, vector<double>& probabilitiesVec, vector<double>& pheromonesVec) {
    double pheromoneSum = 0.0;
    int n = candidateCities.size();
    probabilitiesVec.resize(n);
    for (int i = 0; i < n; ++i) {
        int city = candidateCities[i];
        double pheromone = pow(pheromonesVec[currentCity * cities.size() + city], ALPHA);
        double heuristic = pow(1.0 / distance(cities[currentCity], cities[city]), BETA);
        probabilitiesVec[i] = pheromone * heuristic;
        pheromoneSum += probabilitiesVec[i];
    }
    for (int i = 0; i < n; ++i) {
        probabilitiesVec[i] /= pheromoneSum;
    }
}

// 选择下一个城市
int selectNextCity(int currentCity, vector<int>& candidateCities, vector<City>& cities, vector<double>& pheromonesVec) {
    vector<double> probabilitiesVec;
    probabilities(currentCity, candidateCities, cities, probabilitiesVec, pheromonesVec);
    double r = static_cast<double>(rand()) / static_cast<double>(RAND_MAX);
    double sum = 0.0;
    int n = candidateCities.size();
    for (int i = 0; i < n; ++i) {
        sum += probabilitiesVec[i];
        if (r <= sum) {
            return candidateCities[i];
        }
    }
    return candidateCities.back();
}

// 蚂蚁进行一次迭代
void antTour(Ant& ant, vector<City>& cities, vector<double>& pheromonesVec) {
    int n = cities.size();
    int startCity = rand() % n;
    ant.tour.push_back(startCity);
    ant.visited[startCity] = true;
    for (int i = 0; i < n - 1; ++i) {
        int currentCity = ant.tour.back();
        vector<int> candidateCities;
        for (int j = 0; j < n; ++j) {
            if (!ant.visited[j]) {
                candidateCities.push_back(j);
            }
        }
        int nextCity = selectNextCity(currentCity, candidateCities, cities, pheromonesVec);
        ant.tour.push_back(nextCity);
        ant.visited[nextCity] = true;
        ant.length += distance(cities[currentCity], cities[nextCity]);
    }
    ant.length += distance(cities[ant.tour.back()], cities[startCity]);
}

// 更新信息素
void updatePheromones(vector<Ant>& ants, vector<City>& cities, vector<double>& pheromonesVec) {
    int n = cities.size();
    int m = ants.size();
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            double delta = 0.0;
            for (int k = 0; k < m; ++k) {
                int p = ants[k].tour[i];
                int q = ants[k].tour[j];
                if (p == i && q == j) {
                    delta += Q / ants[k].length;
                }
            }
            pheromonesVec[i * n + j] = pheromonesVec[i * n + j] * (1.0 - RHO) + delta;
            pheromonesVec[j * n + i] = pheromonesVec[i * n + j];
        }
    }
}

// 蚁群算法
vector<int> antColonyOptimization(vector<City>& cities, int numAnts) {
    vector<int> bestTour;
    double bestLength = MAX_DISTANCE;
    int n = cities.size();
    vector<double> pheromonesVec(n * n, 0.01);
    vector<Ant> ants(numAnts, Ant(n));
    for (int iter = 0; iter < MAX_ITERATIONS; ++iter) {
        for (int k = 0; k < numAnts; ++k) {
            ants[k].tour.clear();
            ants[k].visited.assign(n, false);
            ants[k].length = 0.0;
            antTour(ants[k], cities, pheromonesVec);
            if (ants[k].length < bestLength) {
                bestLength = ants[k].length;
                bestTour = ants[k].tour;
            }
        }
        updatePheromones(ants, cities, pheromonesVec);
    }
    return bestTour;
}

// 主函数
int main() {
    srand(static_cast<unsigned int>(time(0)));
    vector<City> cities;
    cities.push_back(City(0, 60.0, 200.0));
    cities.push_back(City(1, 180.0, 200.0));
    cities.push_back(City(2, 80.0, 180.0));
    cities.push_back(City(3, 140.0, 180.0));
    cities.push_back(City(4, 20.0, 160.0));
    cities.push_back(City(5, 100.0, 160.0));
    cities.push_back(City(6, 200.0, 160.0));
    cities.push_back(City(7, 140.0, 140.0));
    cities.push_back(City(8, 40.0, 120.0));
    cities.push_back(City(9, 100.0, 120.0));
    cities.push_back(City(10, 180.0, 100.0));
    cities.push_back(City(11, 60.0, 80.0));
    cities.push_back(City(12, 120.0, 80.0));
    cities.push_back(City(13, 180.0, 60.0));
    cities.push_back(City(14, 20.0, 40.0));
    cities.push_back(City(15, 100.0, 40.0));
    cities.push_back(City(16, 200.0, 40.0));
    cities.push_back(City(17, 20.0, 20.0));
    cities.push_back(City(18, 60.0, 20.0));
    cities.push_back(City(19, 160.0, 20.0));
    vector<int> tour = antColonyOptimization(cities, 50);
    double length = pathLength(tour, cities);
    cout << "Tour length: " << length << endl;
    cout << "Tour: ";
    for (int i = 0; i < tour.size(); ++i) {
        cout << tour[i] << " ";
    }
    cout << endl;
    return 0;
}

该代码实现了一个简单的蚁群算法来求解旅行商问题。其中，cities数组存储城市信息，pheromonesVec数组存储信息素信息，Ant结构体表示蚂蚁，antColonyOptimization函数实现了蚁群算法的主要逻辑，pathLength函数用于计算一条路径的长度。您可以根据自己的需求进行修改和扩展。