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原文 How many planets are out there? Today scientists believe that planets could outnumber the stars.For centuries,scientists and natual philosophers have proposed that stars in the night sky have planetary system similar to our solar system.The existe…
[SIGGRAPH 2015][巫师3 狂猎 The Witcher 3: Wild Hunt ]顶级的开放世界游戏的实现技术 作者:西川善司 日文链接  http://www.4gamer.net/games/202/G020288/20150811091/       计算机图形和交互技术的学术大会[SIGGRAPH 2015],在北美时间的8月9日到13日召开了.            SIGGRAPH 2015的会场,因E3而被熟知的洛杉矶会议中心        SIGGRAPH有着美国…
Treasure Hunt IV Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is exploring the wonderland, suddenly she fell into a hole, when she woke up, she found there are b - a + 1 treasures labled a from b in front of her.Alice was very excited but…
Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4797   Accepted: 1998 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-ar…
其中,将洞穴连起来的算法要好好体会. 学习构建临时变量列表,确认循环用FOR,非确定循环用 WHILE,并定好退出条件. from random import choice cave_numbers = range(0,20) caves = [] for i in cave_numbers: caves.append([]) #保证所有洞穴双向连通 unvisited_caves = range(0,20) visited_caves = [0] unvisited_caves.remove(…
有空就要慢慢练起~~~~脑袋动起来是很快乐的事儿....:) <易学PYTHON>演练一遍. from random import choice cave_numbers = range(1,21) wumpus_location = choice(cave_numbers) player_location = choice(cave_numbers) while player_location == wumpus_location: player_location = choice(cave…
昨天看到微软出的网游Code Hunt.o(∩_∩)o...哈哈,还不好好玩一吧,个人感觉不是一个模块比一个模块难的,Code Hunt是按功能划分.所以不要怕自己做不来.由于不同人特长不一样. 像ACM都是分工合作的啦.废话不多.我们来总结一下前01-04的答案.希望对大家有帮助,纯属娱乐.还有非常多能够优化的地方,欢迎一起来讨论! 注:语言C# 网页地址:https://www.codehunt.com/ 转载请著明:http://blog.csdn.net/grublinux/articl…
Treasure Hunt IV Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice is exploring the wonderland, suddenly she fell into a hole, when she woke up, she found there are b - a + 1 treasures labled a from b in front of her. Alice was very excited but…
Treasure Hunt 大意:在一个矩形区域内.有n条线段,线段的端点是在矩形边上的,有一个特殊点,问从这个点到矩形边的最少经过的线段条数最少的书目,穿越仅仅能在中点穿越. 思路:须要巧妙的转换一下这个问题,由于从一个点到终点不可能"绕过"围墙.仅仅能穿过去,所以门是否开在中点是无所谓的,仅仅要求四周线段中点到终点的线段与墙的最少交点个数就可以.更进一步,实际上,仅仅需推断四周围墙的全部点与终点的连线与内墙的最少交点加一就可以. struct Point{ double x, y;…
//开学了,好烦啊啊啊啊啊!怎么开个学那么多破事情!!都俩星期了,终于有时间写出来一道题 题意:不难理解,不写了.这几天忙的心累. 代码:(Accepted, 0.010s) //UVa1596 - Bug Hunt #include<iostream> #include<sstream> #include<string> #include<stack> #include<map> using namespace std; struct o_O…