C++程序实现多机器人cbs算法

CBS算法是一种针对多机器人路径规划问题的算法，其核心思想是将多机器人的路径规划问题分解为单机器人的路径规划问题，并利用冲突检测和冲突消解来解决整个问题。下面是一个基于C++语言实现的多机器人CBS算法的示例代码：

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

// 定义机器人结构体
struct Robot {
    int id;         // 机器人编号
    int start;      // 起点编号
    int goal;       // 终点编号
    vector<int> path;   // 路径
};

// 定义冲突结构体
struct Conflict {
    int time;       // 冲突时间
    int robot1;     // 机器人1编号
    int robot2;     // 机器人2编号
    int location;   // 冲突位置
};

// 定义节点结构体
struct Node {
    vector<Robot> robots;   // 机器人集合
    vector<Conflict> conflicts; // 冲突集合
    int cost;       // 路径长度
    bool operator<(const Node &n) const {
        return cost > n.cost;
    }
};

// 定义地图
vector<vector<int>> map;

// 定义机器人集合和终点集合
vector<Robot> robots;
vector<int> goals;

// 定义冲突检测函数
bool detect_conflict(int time, int robot1, int robot2, vector<Robot> &robots) {
    int loc1 = robots[robot1].path[time];
    int loc2 = robots[robot2].path[time];
    if (loc1 == loc2) {
        return true;
    }
    for (int i = time + 1; i < robots[0].path.size(); i++) {
        int next_loc1 = robots[robot1].path[i];
        int next_loc2 = robots[robot2].path[i];
        if (next_loc1 == next_loc2) {
            return true;
        }
        if (next_loc1 == loc2 && next_loc2 == loc1) {
            return true;
        }
        loc1 = next_loc1;
        loc2 = next_loc2;
    }
    return false;
}

// 定义冲突消解函数
void resolve_conflict(int time, int robot1, int robot2, vector<Robot> &robots) {
    Robot &r1 = robots[robot1];
    Robot &r2 = robots[robot2];
    int loc1 = r1.path[time];
    int loc2 = r2.path[time];
    if (loc1 == loc2) {
        return;
    }
    // 交换位置
    if (r1.goal == loc2 && r2.goal == loc1) {
        swap(r1.goal, r2.goal);
        swap(r1.path, r2.path);
        return;
    }
    // 机器人1等待
    if (r1.goal == loc2) {
        r1.path.insert(r1.path.begin() + time, loc1);
        return;
    }
    // 机器人2等待
    if (r2.goal == loc1) {
        r2.path.insert(r2.path.begin() + time, loc2);
        return;
    }
    // 交叉
    for (int i = time + 1; i < r1.path.size(); i++) {
        if (r1.path[i] == loc2 && r2.path[i] == loc1) {
            r1.path.insert(r1.path.begin() + i, loc1);
            return;
        }
        if (r2.path[i] == loc1 && r1.path[i] == loc2) {
            r2.path.insert(r2.path.begin() + i, loc2);
            return;
        }
    }
}

// 定义A*算法函数
int astar(int robot, int start, int goal) {
    vector<int> g(map.size(), INT_MAX);
    vector<int> f(map.size(), INT_MAX);
    vector<int> parent(map.size(), -1);
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    g[start] = 0;
    f[start] = g[start] + heuristic(start, goal);
    pq.push({f[start], start});
    while (!pq.empty()) {
        int curr = pq.top().second;
        pq.pop();
        if (curr == goal) {
            int cost = 0;
            while (parent[curr] != -1) {
                int prev = parent[curr];
                cost += distance(curr, prev);
                curr = prev;
            }
            return cost;
        }
        for (int i = 0; i < map[curr].size(); i++) {
            int next = map[curr][i];
            int dist = distance(curr, next);
            int h = heuristic(next, goal);
            if (g[curr] + dist < g[next]) {
                g[next] = g[curr] + dist;
                f[next] = g[next] + h;
                parent[next] = curr;
                pq.push({f[next], next});
            }
        }
    }
    return INT_MAX;
}

// 定义CBS算法
int cbs() {
    Node n;
    n.robots = robots;
    for (int i = 0; i < robots.size(); i++) {
        n.cost += astar(i, robots[i].start, robots[i].goal);
    }
    priority_queue<Node> pq;
    pq.push(n);
    while (!pq.empty()) {
        Node curr = pq.top();
        pq.pop();
        if (curr.conflicts.empty()) {
            return curr.cost;
        }
        Conflict c = curr.conflicts[0];
        vector<Node> children;
        for (int i = 0; i < 2; i++) {
            Node child = curr;
            int robot = (i == 0 ? c.robot1 : c.robot2);
            int goal = (i == 0 ? robots[c.robot1].goal : robots[c.robot2].goal);
            for (int j = c.time; j < child.robots[0].path.size() && child.robots[robot].path[j] != goal; j++) {
                child.robots[robot].path[j] = child.robots[robot].path[c.time];
            }
            child.robots[robot].path.erase(child.robots[robot].path.begin() + c.time + 1, child.robots[robot].path.end());
            child.robots[robot].goal = goal;
            child.cost = 0;
            for (int j = 0; j < robots.size(); j++) {
                child.cost += astar(j, child.robots[j].start, child.robots[j].goal);
            }
            children.push_back(child);
        }
        for (int i = 0; i < children.size(); i++) {
            bool valid = true;
            for (int j = 0; j < children[i].robots.size(); j++) {
                for (int k = j + 1; k < children[i].robots.size(); k++) {
                    if (detect_conflict(c.time, j, k, children[i].robots)) {
                        valid = false;
                        Conflict new_c = {c.time, j, k, children[i].robots[j].path[c.time]};
                        children[i].conflicts.push_back(new_c);
                    }
                }
            }
            if (valid) {
                pq.push(children[i]);
            }
        }
    }
    return INT_MAX;
}

int main() {
    // 初始化地图
    map = vector<vector<int>>(10, vector<int>(10, 1));
    // 初始化机器人和终点
    Robot r1 = {0, 1, 9, {1}};
    Robot r2 = {1, 9, 1, {9}};
    Robot r3 = {2, 1, 9, {1}};
    Robot r4 = {3, 9, 1, {9}};
    robots.push_back(r1);
    robots.push_back(r2);
    robots.push_back(r3);
    robots.push_back(r4);
    goals = {9, 1, 9, 1};
    // 调用CBS算法
    int cost = cbs();
    cout << "Total cost: " << cost << endl;
    return 0;
}

在上述代码中，我们首先定义了机器人和冲突的结构体，然后定义了冲突检测和冲突消解的函数，以及A*算法的函数。最后，我们定义了CBS算法的函数，并在主函数中调用该函数来解决多机器人路径规划问题。