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城市:天际线Cities: Skylines for Mac(城市建造类游戏)+dlcv1.16.0-f3 中文原生版
《城市：天际线》是一款模拟经营类游戏，该游戏可以让玩家设计、建造和管理自己的城市，并在满足居民需求的前提下获得收益。
1.7K30编辑于 2023-04-18 - 来自专栏calmound
Connect the Cities
水题，但是g++超时，c++对了 不解 #include<stdio.h> const int MAXN=505; const int INF=0x7fffffff; int xx[MAXN]; int pre[MAXN]; int map[MAXN][MAXN]; int dist[MAXN]; int ans,n,flag; void Prim() { int i,j,k; int mn; bool p[MAXN]; for(i=2;i<=n;i++) {
782110发布于 2018-04-17 - 来自专栏算法修养
CodeForces 665A Buses Between Cities
Buses Between Cities time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Buses run between the cities A and B, the first one is at Note that you should not count the encounters in cities Aand B.
71440发布于 2018-04-26 - 来自专栏饶文津的专栏
【HDU 2874】Connections between cities（LCA）
i和j在一棵树上，则最短路为dis[i]+dis[j]-dis[LCA(i,j)]*2。
50610发布于 2020-06-02 - 来自专栏ml
HDUOJ---3371Connect the Cities
Connect the Cities Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others Though some survived cities are still connected with others, but most of them become disconnected. k <= 100) stands for the number of still connected cities. To make it easy, the cities are signed from 1 to n. Then t integers follow stands for the id of these cities.
75770发布于 2018-03-22 - 来自专栏hotarugaliの技术分享
PAT「1001 Battle Over Cities - Hard Version (35分)」
题目 题目链接：PAT「1001 Battle Over Cities - Hard Version (35分)」 。 Description It is vitally important to have all the cities connected by highways in a war. To keep the rest of the cities connected, we must repair some highways with the minimum cost. Given the map of cities which have all the destroyed and remaining highways marked, you are supposed Each case starts with a line containing numbers ( , and which are the total number of cities
39420编辑于 2022-03-01 【PAT】甲级1013 - Battle Over Cities（并查集）
题目链接：点击打开题目 Battle Over Cities (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. , the number of remaining highways, and the number of cities to be checked, respectively. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.
23610编辑于 2025-08-27- 来自专栏算法修养
PAT 1013 Battle Over Cities(并查集)
Battle Over Cities (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN , Yue It is vitally important to have all the cities connected by highways in a war. For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.
78560发布于 2018-04-27 Cradle：颠覆AI Agent 操作本地软件，AI驱动的通用计算机控制框架，如何让基础模型像人一样操作你的电脑？
实验证明，Cradle 可完成：AAA 游戏：Red Dead Redemption 2 主线任务、高成功率完成；市政游戏：Cities: Skylines 打造千人城市；农场游戏：Stardew Valley
1.3K10编辑于 2025-08-01- 来自专栏VRPinea
VR版《捉鬼敢死队》来袭，今年的Meta Quest游戏展示会还有哪些惊喜？
Cities: VR 开发商：Fast Travel Games 上线时间：4月28日 平台：Quest 2 是知名模拟建造游戏《城市：天际线（Cities: Skylines）》的衍生作品，且即将发布
76220编辑于 2022-06-08 - 来自专栏vblog
PAT 1013 Battle Over Cities (25分) 图的连通分量+DFS
题目 It is vitally important to have all the cities connected by highways in a war. For example, if we have 3 cities and 2 highways connecting city1 -city2 and city1 -city3 , the number of remaining highways, and the number of cities to be checked, respectively. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.
41010发布于 2020-07-14 - 来自专栏机器学习从入门到成神
2017美国数学建模ICM E题 可持续发展的城市需要（Sustainable Cities Needed!）
背景： 许多社区正在实施智能增长计划，以考虑长期，可持续的规划目标。 “智能增长是要帮助每个城镇和城市，使它们变成更加经济繁荣，社会公平和环境可持续的生活场所。”智能增长的重点是建设可持续发展的城市 - 经济繁荣，社会公平，环境可持续。这个任务比以往任何时候都重要，因为世界正在迅速城市化。预计到2050年， 66％的世界人口将是城市人口 - 这将导致25亿人口被纳入城市人口。因此，城市规划变得越来越重要和必要，以确保人们获得公平和可持续的家园，资源和就业机会。
1K10发布于 2018-09-14 - 来自专栏数据魔术师
干货 | 变邻域搜索算法(VNS)求解TSP（附C++详细代码及注释）
_2city(cities[tmp[i - 1]], cities[tmp[k]]) 155 - distance_2city(cities[tmp[k]], cities [cities_permutation[k]], cities[cities_permutation[i+1]]) 289 + distance_2city(cities[ [k]]) 299 + distance_2city(cities[cities_permutation[k]], cities[cities_permutation[i [cities_permutation[0]], cities[cities_permutation[k]]) 313 - distance_2city(cities[cities_permutation [k]], cities[cities_permutation[k-1]]) 314 + distance_2city(cities[cities_permutation[
4.3K20发布于 2018-07-31 - 来自专栏数据魔术师
干货|迭代局部搜索算法(Iterated local search)探幽（附C++代码及注释）
* cities_permutation, CITIES * cities); //获取随机城市排列, 用于产生初始解 void random_permutation(int * cities_permutation int i, int k, int *tmp, CITIES * cities); //城市排列 int permutation[CITY_SIZE]; //城市坐标数组 CITIES cities [i]; cities_permutation[i] = cities_permutation[r]; cities_permutation[r] = temp; (cities[tmp[0]], cities[tmp[k]]) + distance_2city(cities[tmp[i]], cities[tmp[0]]); _2city(cities[tmp[k]], cities[tmp[k + 1]]) + distance_2city(cities[tmp[i]], cities[tmp
3.9K90发布于 2018-06-11 - 来自专栏程序猿声
【优化算法】迭代局部搜索算法(Iterated local search)探幽（附C++代码及注释）
(int *cities_permutation, int *old_cities_permutation, CITIES * p_cities); 65 66//将城市序列分成4块，然后按块重新打乱顺序 [i]; 174 cities_permutation[i] = cities_permutation[r]; 175 cities_permutation[r] = temp [tmp[k]], cities[tmp[k + 1]]) 233 + distance_2city(cities[tmp[i]], cities[tmp[k + 1]]) [tmp[i - 1]], cities[tmp[i]]) 245 + distance_2city(cities[tmp[i - 1]], cities[tmp[k]]) _2city(cities[tmp[i - 1]], cities[tmp[k]]) 254 - distance_2city(cities[tmp[k]], cities
2.4K11发布于 2019-05-14 - 来自专栏数据魔术师
变邻域搜索算法(VNS)求解TSP（附Java详细代码及注释）
2city(cities[cities_permutation[i]], cities[cities_permutation[i + 1]]) - distance_2city(cities [k]], cities[cities_permutation[k + 1]]) + distance_2city(cities[cities_permutation[k - 1]], cities [0]], cities[cities_permutation[k]]) - distance_2city(cities[cities_permutation[k]], cities[cities_permutation (cities[cities_permutation[i]], cities[cities_permutation[i + 1]]) - distance_2city(cities[cities_permutation [k]], cities[cities_permutation[k + 1]]) - distance_2city(cities[cities_permutation[k]], cities
2.1K41发布于 2021-09-02 - 来自专栏数据科学CLUB
蚁群算法
构建城市列表 self.city_dist_mat = np.zeros((self.cities_num, self.cities_num)) # 3 构建城市距离矩阵 for i in range(self.cities_num): A = self.cities[i] for j in range(i, self.cities_num): B = self.cities[j] self.city_dist_mat[i][ ] = np.random.permutation(range(self.cities_num)) temp_index = self.cities_num + self.cities_num] = np.random.permutation(range(self.cities_num)) temp_index += self.cities_num
1.5K40发布于 2020-06-12 - 来自专栏前端砍柴人
怒肝 JavaScript 数据结构 — 数组篇(二)
假设现在有一个数组 cities 如下： var cities = ['北京', '上海', '杭州', '深圳'] 我们要通过遍历数组，每个数组项前面加上 中国- 这个前缀，用基本的 for 循环实现如下 ： for(var i = 0; i < cities.length; i++) { cities[i] = '中国-' + cities[i]; } // cities = ['中国-北京', 比如上面的循环可以用 forEach 替代： cities.forEach((item, i)=> { cities[i] = '中国-' + item; }) // cities = ['中国 下面几个迭代器的参数也是这个回调函数： map filter find findIndex some every 我们再用 map 实现上面的逻辑： cities = cities.map(item=> 比如将数组 cities 中的第三和第四个数组项替换成 红旗，实现如下： cities.fill('红旗', 2, 4); // cities：['北京', '上海', '红旗', '红旗'] 注意：
1.4K41编辑于 2022-09-22 - 来自专栏数据结构和算法
Python高级算法——模拟退火算法（Simulated Annealing）
): total = 0 for i in range(len(order) - 1): total += distance(cities[order[i]], cities [order[i + 1]]) return total + distance(cities[order[-1]], cities[order[0]]) def simulated_annealing cooling_rate return best_order # 示例 np.random.seed(42) num_cities = 10 cities = np.random.rand( num_cities, 2) initial_order = np.arange(num_cities) np.random.shuffle(initial_order) final_order = ) print("最优解的总距离:", total_distance(final_order, cities)) 应用场景 5.
3.4K10编辑于 2023-12-18 - 来自专栏热度文章
杨贵妃的荔枝之旅：用Dijkstra算法穿越中国
数据结构设计采用邻接矩阵表示城市网络：double graph[NUM_CITIES][NUM_CITIES];这种结构虽然空间复杂度较高(O(V²))，但对于中等规模城市网络(V=10)完全可接受，且具有以下优势 NUM_CITIES] = {0}; // 已访问标记 int prev[NUM_CITIES]; // 前驱节点 int path[NUM_CITIES]; city_names[index] : "未知";}// Dijkstra 算法实现void dijkstra(double graph[NUM_CITIES][NUM_CITIES], int start , int end) { double dist[NUM_CITIES]; // 最短距离数组 int visited[NUM_CITIES] = {0}; // 已访问标记 ][NUM_CITIES]; // 初始化所有节点间距离为无穷大 for (int i = 0; i < NUM_CITIES; i++) { for (int j =
25721编辑于 2025-06-21
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