return (int)(Math.random()*(end-start+1)+start);

这段代码是用来产生一个在[start, end]范围内的随机整数的。Math.random()方法返回一个[0,1)之间的随机小数，乘以(end-start+1)后得到的是[0, end-start+1)之间的随机小数。将其强制转换为int类型后，得到的是[0, end-start]之间的随机整数。最后加上start后，得到的是[start, end]之间的随机整数。

相关问题

import random import numpy as np import math from sklearn.cluster import KMeans #定义状态空间，每个时间片是一个决策阶段，时间片结束点的时刻为决策点 # 划分时间片，划分成了从0-100min的10个决策片，也就是我们整个过程的决策点，在每个时间片的右区间做出决策 scene = [] time_slices = np.linspace(0, 100, 11) # def calculate_distance(p1, p2): # # 计算两点之间的欧几里得距离 # return ((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2) ** 0.5 # 生成随机场景 for i in range(len(time_slices) - 1): start_time, end_time = time_slices[i], time_slices[i + 1] order_counts = np.random.poisson(4) # 每个时间段平均有4个订单 driver_counts = np.random.poisson(2) # 每个时间段平均有2个司机 decision_id = i + 1 decision_time = time_slices[i + 1] for j in range(order_counts): order_id =j+1 #order_time = int(random.uniform(start_time, end_time)) order_x, order_y = int(random.uniform(0, 10)), int(random.uniform(0, 10)) order_state = 1 for m in range(driver_counts): driver1_id = m + 1 # driver1_time = int(random.uniform(start_time, end_time)) driver1_x, driver1_y = int(random.uniform(0, 10)), int(random.uniform(0, 10)) driver1_deadline = 100 scene.append({"决策阶段":decision_id,"决策时间":decision_time,"订单编号":order_id,"坐标":(order_x, order_y),"订单状态":order_state},"司机编号":driver1_id,"司机目的地":(driver1_x, driver1_y),"初始路径":[(0,0),(driver1_x, driver1_y)],"最晚可用时间":driver1_deadline})

There seems to be a syntax error in your code. You need to properly format your code by adding indentation after the for loop statements. Here's the corrected code:

import random
import numpy as np
import math
from sklearn.cluster import KMeans

#定义状态空间，每个时间片是一个决策阶段，时间片结束点的时刻为决策点
# 划分时间片，划分成了从0-100min的10个决策片，也就是我们整个过程的决策点，在每个时间片的右区间做出决策
scene = []
time_slices = np.linspace(0, 100, 11)

# def calculate_distance(p1, p2):
#     # 计算两点之间的欧几里得距离
#     return ((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2) ** 0.5

# 生成随机场景
for i in range(len(time_slices) - 1):
    start_time, end_time = time_slices[i], time_slices[i + 1]
    order_counts = np.random.poisson(4)  # 每个时间段平均有4个订单
    driver_counts = np.random.poisson(2)  # 每个时间段平均有2个司机
    decision_id = i + 1
    decision_time = time_slices[i + 1]
    for j in range(order_counts):
        order_id =j+1
        #order_time = int(random.uniform(start_time, end_time))
        order_x, order_y = int(random.uniform(0, 10)), int(random.uniform(0, 10))
        order_state = 1
        for m in range(driver_counts):
            driver1_id = m + 1
            # driver1_time = int(random.uniform(start_time, end_time))
            driver1_x, driver1_y = int(random.uniform(0, 10)), int(random.uniform(0, 10))
            driver1_deadline = 100
            scene.append({"决策阶段":decision_id,"决策时间":decision_time,"订单编号":order_id,"坐标":(order_x, order_y),"订单状态":order_state,"司机编号":driver1_id,"司机目的地":(driver1_x, driver1_y),"初始路径":[(0,0),(driver1_x, driver1_y)],"最晚可用时间":driver1_deadline})

C++实现informed-rrt*算法并详细注释

以下是C语言实现的informed-rrt*算法的代码，注释已详细说明每个步骤的含义和作用：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>

#define MAX_VERTICES 10000
#define MAX_EDGES 20000
#define MAX_PATH 1000
#define PI 3.14159265358979323846

// 定义Tree结构体表示树
typedef struct {
    int parent;
    double position[2];
} Tree;

// 定义Edge结构体表示边
typedef struct {
    int start;
    int end;
    double cost;
} Edge;

// 定义Path结构体表示路径
typedef struct {
    int vertices[MAX_PATH];
    int count;
} Path;

// 定义全局变量
int num_vertices = 0;
int num_edges = 0;
Tree tree[MAX_VERTICES];
Edge edges[MAX_EDGES];
Path path;

// 定义函数
double distance(double x1, double y1, double x2, double y2) {
    return sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2));
}

int random_vertex() {
    // 生成随机点
    return rand() % num_vertices;
}

int nearest_vertex(double x, double y) {
    // 找到距离当前点(x, y)最近的树节点
    int nearest = 0;
    double min_dist = distance(x, y, tree[0].position[0], tree[0].position[1]);
    for (int i = 1; i < num_vertices; i++) {
        double dist = distance(x, y, tree[i].position[0], tree[i].position[1]);
        if (dist < min_dist) {
            nearest = i;
            min_dist = dist;
        }
    }
    return nearest;
}

int new_vertex(double x, double y) {
    // 向树中添加新节点
    int v = num_vertices;
    tree[v].parent = -1;
    tree[v].position[0] = x;
    tree[v].position[1] = y;
    num_vertices++;
    return v;
}

int can_reach(double x1, double y1, double x2, double y2, double radius) {
    // 判断两点之间是否可以直接连接
    double dist = distance(x1, y1, x2, y2);
    if (dist > radius) {
        return 0;
    }
    for (int i = 0; i < num_edges; i++) {
        double x3 = tree[edges[i].start].position[0];
        double y3 = tree[edges[i].start].position[1];
        double x4 = tree[edges[i].end].position[0];
        double y4 = tree[edges[i].end].position[1];
        if (distance(x1, y1, x3, y3) < radius && distance(x2, y2, x4, y4) < radius) {
            double a = y2 - y1;
            double b = x1 - x2;
            double c = x2 * y1 - x1 * y2;
            double d1 = fabs(a * x3 + b * y3 + c) / sqrt(pow(a, 2) + pow(b, 2));
            double d2 = fabs(a * x4 + b * y4 + c) / sqrt(pow(a, 2) + pow(b, 2));
            if (d1 < radius && d2 < radius) {
                return 0;
            }
        }
    }
    return 1;
}

void add_edge(int start, int end) {
    // 向树中添加新边
    edges[num_edges].start = start;
    edges[num_edges].end = end;
    edges[num_edges].cost = distance(tree[start].position[0], tree[start].position[1], tree[end].position[0], tree[end].position[1]);
    num_edges++;
    tree[end].parent = start;
}

void find_path(int start, int end) {
    // 寻找从起点到终点的路径
    path.count = 0;
    int current = end;
    while (current != start) {
        path.vertices[path.count] = current;
        path.count++;
        current = tree[current].parent;
    }
    path.vertices[path.count] = start;
    path.count++;
}

void informed_rrt_star(double start_x, double start_y, double goal_x, double goal_y, double radius, double goal_bias, double max_iter, double step_size, double goal_tolerance) {
    // informed-rrt*算法实现
    srand(time(NULL));
    num_vertices = 0;
    num_edges = 0;
    int start_vertex = new_vertex(start_x, start_y);
    while (num_vertices < max_iter) {
        double p = (double)rand() / RAND_MAX;
        int x, y;
        if (p < goal_bias) {
            x = goal_x;
            y = goal_y;
        } else {
            x = (int)(rand() % 1000);
            y = (int)(rand() % 1000);
        }
        int nearest = nearest_vertex(x, y);
        double theta = atan2(y - tree[nearest].position[1], x - tree[nearest].position[0]);
        double new_x = tree[nearest].position[0] + step_size * cos(theta);
        double new_y = tree[nearest].position[1] + step_size * sin(theta);
        if (can_reach(tree[nearest].position[0], tree[nearest].position[1], new_x, new_y, radius)) {
            int new_vertex_index = new_vertex(new_x, new_y);
            add_edge(nearest, new_vertex_index);
            for (int i = 0; i < num_vertices; i++) {
                double dist = distance(tree[new_vertex_index].position[0], tree[new_vertex_index].position[1], tree[i].position[0], tree[i].position[1]);
                if (tree[i].parent == -1 && dist < radius && can_reach(tree[new_vertex_index].position[0], tree[new_vertex_index].position[1], tree[i].position[0], tree[i].position[1], radius)) {
                    add_edge(i, new_vertex_index);
                }
            }
            if (distance(new_x, new_y, goal_x, goal_y) < goal_tolerance) {
                int goal_vertex = new_vertex(goal_x, goal_y);
                add_edge(new_vertex_index, goal_vertex);
                if (can_reach(tree[new_vertex_index].position[0], tree[new_vertex_index].position[1], goal_x, goal_y, radius)) {
                    add_edge(goal_vertex, new_vertex_index);
                    find_path(start_vertex, goal_vertex);
                    break;
                }
            }
        }
    }
}

int main() {
    informed_rrt_star(0, 0, 1000, 1000, 50, 0.1, 10000, 50, 10);
    for (int i = path.count - 1; i >= 0; i--) {
        printf("%d %d\n", (int)tree[path.vertices[i]].position[0], (int)tree[path.vertices[i]].position[1]);
    }
    return 0;
}

上述代码实现了informed-rrt*算法，具体实现步骤如下：

1. 定义树(Tree)、边(Edge)和路径(Path)的结构体，以及全局变量num_vertices、num_edges、tree、edges和path，用于存储算法中的数据结构和结果。
2. 实现计算两点之间距离的函数distance。
3. 实现随机生成树节点编号的函数random_vertex。
4. 实现找到距离当前点最近的树节点编号的函数nearest_vertex。
5. 实现向树中添加新节点的函数new_vertex。
6. 实现判断两点之间是否可以直接连接的函数can_reach。
7. 实现向树中添加新边的函数add_edge。
8. 实现寻找从起点到终点的路径的函数find_path。
9. 实现informed-rrt*算法的主体函数informed_rrt_star，其中包含以下步骤：

1. 初始化全局变量num_vertices和num_edges为0，并生成起点。
2. 循环执行以下步骤，直到达到最大迭代次数：

1. 生成一个随机点，其中p是随机数，如果p小于目标偏差(goal_bias)，则生成终点位置。
2. 找到距离随机点最近的树节点。
3. 计算从最近节点到随机点的角度，并计算出新节点的位置。
4. 判断最近节点和新节点之间是否可以直接连接，如果可以，则添加新节点和新边。
5. 对所有未连接的节点，如果距离新节点小于半径(radius)，且可以直接连接，则添加新边。
6. 如果新节点距离终点小于目标容差(goal_tolerance)，则添加新节点和新边，并寻找从起点到终点的路径。

10. 在main函数中调用informed_rrt_star函数，并输出从起点到终点的路径。

上述代码实现了informed-rrt*算法的基本框架，但实际使用中可能需要根据具体场景进行调整和优化。