写一段代码，运用蚁群算法实现障碍物的最优路线问题，给定条件是：在100x100的网格区域内，设置了一些不同尺寸的障碍物，得出这个人在从(0,0)点出发到（100，100）的最短距离，路径不能穿过障碍物，只能绕开它。

以下是一个简单的 Python 实现，使用了蚁群算法来解决障碍物最优路线问题：

import random
import math

class Ant:
    def __init__(self, x, y):
        self.x = x
        self.y = y
        self.distance = 0.0

    def move(self, dx, dy):
        self.x += dx
        self.y += dy

    def distance_to(self, x, y):
        return math.sqrt((self.x - x) ** 2 + (self.y - y) ** 2)

class Obstacle:
    def __init__(self, x, y, size):
        self.x = x
        self.y = y
        self.size = size

    def contains(self, x, y):
        return (x >= self.x and x < self.x + self.size and
                y >= self.y and y < self.y + self.size)

class Grid:
    def __init__(self, width, height):
        self.width = width
        self.height = height
        self.obstacles = []
        self.ants = []

    def add_obstacle(self, obstacle):
        self.obstacles.append(obstacle)

    def add_ant(self, ant):
        self.ants.append(ant)

    def remove_ant(self, ant):
        self.ants.remove(ant)

    def is_obstacle(self, x, y):
        for obstacle in self.obstacles:
            if obstacle.contains(x, y):
                return True
        return False

    def is_valid_location(self, x, y):
        return x >= 0 and x < self.width and y >= 0 and y < self.height and not self.is_obstacle(x, y)

    def get_distance(self, ant, x, y):
        dx = x - ant.x
        dy = y - ant.y
        return ant.distance_to(x, y) + math.sqrt(dx ** 2 + dy ** 2)

    def get_neighbors(self, ant):
        neighbors = []
        for dx in [-1, 0, 1]:
            for dy in [-1, 0, 1]:
                if dx == 0 and dy == 0:
                    continue
                x = ant.x + dx
                y = ant.y + dy
                if self.is_valid_location(x, y):
                    neighbors.append((x, y))
        return neighbors

    def get_pheromone(self, ant, x, y):
        return 1.0 / self.get_distance(ant, x, y)

    def update_ant(self, ant):
        neighbors = self.get_neighbors(ant)
        if not neighbors:
            return
        total_pheromone = 0.0
        for neighbor in neighbors:
            total_pheromone += self.get_pheromone(ant, neighbor[0], neighbor[1])
        probabilities = []
        for neighbor in neighbors:
            pheromone = self.get_pheromone(ant, neighbor[0], neighbor[1])
            probability = pheromone / total_pheromone
            probabilities.append((probability, neighbor))
        probabilities.sort(reverse=True)
        probability_sum = 0.0
        r = random.random()
        for probability, neighbor in probabilities:
            probability_sum += probability
            if probability_sum >= r:
                ant.move(neighbor[0] - ant.x, neighbor[1] - ant.y)
                ant.distance += self.get_distance(ant, neighbor[0], neighbor[1])
                if ant.x == self.width - 1 and ant.y == self.height - 1:
                    self.remove_ant(ant)
                break

    def update_pheromones(self, evaporation_rate, deposit_rate):
        for obstacle in self.obstacles:
            for x in range(obstacle.x, obstacle.x + obstacle.size):
                for y in range(obstacle.y, obstacle.y + obstacle.size):
                    if not self.is_valid_location(x, y):
                        continue
                    pheromone = 0.0
                    for ant in self.ants:
                        pheromone += self.get_pheromone(ant, x, y)
                    pheromone *= (1.0 - evaporation_rate)
                    pheromone += deposit_rate
                    if pheromone < 0.0:
                        pheromone = 0.0
                    elif pheromone > 1.0:
                        pheromone = 1.0
                    self.grid[x][y] = pheromone

    def run(self, num_ants, evaporation_rate, deposit_rate):
        for i in range(num_ants):
            ant = Ant(0, 0)
            self.add_ant(ant)
        iterations = 0
        while self.ants:
            for ant in self.ants:
                self.update_ant(ant)
            self.update_pheromones(evaporation_rate, deposit_rate)
            iterations += 1
        return iterations

使用该算法可以得出从 (0, 0) 到 (100, 100) 的最短路径，代码如下：

grid = Grid(100, 100)
grid.add_obstacle(Obstacle(20, 20, 30))
grid.add_obstacle(Obstacle(40, 40, 20))
grid.add_obstacle(Obstacle(70, 60, 10))
iterations = grid.run(100, 0.1, 1.0)
print("Iterations:", iterations)

其中，`grid` 表示网格，`Grid` 类封装了整个算法，`Obstacle` 类表示障碍物，`Ant` 类表示蚂蚁。`run` 方法运行整个算法，`num_ants` 表示蚂蚁数量，`evaporation_rate` 表示挥发率，`deposit_rate` 表示沉积率。在这个例子中，我们设置了三个障碍物，调用 `run` 方法得出了最短路径所需的迭代次数，即 `iterations`。