题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5299 题意: 在欧几里得平面上有n个圆,圆之间不会相交也不会相切,现在Alice和Bob玩游戏,两人轮流选择一个圆删除它和它包含的所有圆(Alice先手),在自己的轮次无圆可删判输,问你谁会赢得比赛 解:先将圆之间的关系抽象,转化为一个树图,然后套SG定理(WoW,我可不知道什么是SG定理 想详细了解的话,这里是和SG函数相关的东西:http://www.cnblogs.com/shjwudp/ar…

Circles Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1427    Accepted Submission(s): 451 Problem Description There are n circles on a infinitely large table.With every two circle, eithe…

Time Limit: 3000MS     64bit IO Format: %lld & %llu map存人名,floyd传递闭包,DFS查询. 输出答案的逗号后面还有个空格,被坑到了233 /*by SilverN*/ #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<map&g…

zw版[转发·台湾nvp系列Delphi例程]HALCON 3D Position Of Circles procedure TForm1.action();var ho_Image, ho_EllipseContoursLarge : HUntypedObjectX; ho_EllipseContoursSmall, ho_EllipseContours : HUntypedObjectX; hv_Width, hv_Height : OleVariant; hv_WindowHandle,…

题意:给出n个人的m次电话,问最后构成多少个环,找出所有的环 自己想的是:用map来储存人名,每个人名映射成一个数字编号,再用并查集,求出有多少块连通块,输出 可是map不熟,写不出来,而且用并查集输出的时候感觉貌似很麻烦 然后再用的传递闭包,可是判断到d[i][j]==1和d[j][i]==1,该怎么输出路径呢 于是看了lrj的代码= = 用的是一个ID(string s)函数来给名字编号,和第五章的集合栈计算机那题的办法一样 然后用dfs输出路径= =(这个要好好--好好--好好学) 最后还…

题意:两人相互打电话(直接或间接),则在一个电话圈.即a给b打电话,b给c打电话,则a给c间接打电话. 注意:1.注意标记.2.注意输出格式. #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<sstream> #include<cctype> #include<cmath> #include<cstdlib…

题意: 有n个人m通电话,如果有两个人相互打电话(直接或间接)则在同一个电话圈里.输出所有电话圈的人的名单. 分析: 根据打电话的关系,可以建一个有向图,然后用Warshall算法求传递闭包. 最后输出是辅助一个标记数组,用DFS输出的,这个办法挺巧妙的. 本来我原来的想法是,用并查集求所有的连通分量,然后再好多次循环找连通分量一致的名字输出,那样太麻烦了. ios::sync_with_stdio(false);这个最好不要随便用,可能会产生某些副作用. 字符指针是可以传给string对象作参…

There are n circles on a infinitely large table.With every two circle, either one contains another or isolates from the other.They are never crossed nor tangent.Alice and Bob are playing a game concerning these circles.They take turn to play,Alice go…

  He's Circles He wrote n letters "X" and "E" in a circle. He thought that there were 2n possibilities to do it, because each letter may be either "X" or "E". But Qc noticed that some different sequences of letters…

题目链接 题意: 给定一张有向图.找出全部强连通分量,并输出. 思路:有向图的强连通分量用Tarjan算法,然后用map映射,便于输出,注意输出格式. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <map> #include <algorithm> using namespace std; const int MAXN = 2000; const in…