[Cerc2005]Knights of the Round Table】的更多相关文章

题目描述 有n个骑士经常举行圆桌会议,商讨大事.每次圆桌会议至少有3个骑士参加,且相互憎恨的骑士不能坐在圆桌的相邻位置.如果发生意见分歧,则需要举手表决,因此参加会议的骑士数目必须是大于1的奇数,以防止赞同和反对票一样多.知道那些骑士相互憎恨之后,你的任务是统计有多少骑士不可能参加任何一个会议. 输入格式 包含多组数据,每组数据格式如下: 第一行为两个整数n和m(1<=n<=1000, 1<=m<=10^6). 以下m行每行包含两个整数k1和k2(1<=k1,k2<=n…
Knights of the Round Table Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 12439   Accepted: 4126 Description Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the oth…
Knights of the Round Table Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 10911   Accepted: 3587 Description Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the oth…
Knights of the Round Table Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 9169   Accepted: 2960 Description Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the othe…
Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, anddrinking with the other knights are fun things to do. Therefore, it is not very surprising that in recentyears the kingdom of King Arthur has exp…
Description Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the other knights are fun things to do. Therefore, it is not very surprising that in recent years the kingdom of King…
Problem  UVALive - 3523 - Knights of the Round Table Time Limit: 4500 mSec Problem Description Input The input contains several blocks of test cases. Each case begins with a line containing two integers 1 ≤ n ≤ 1000 and 1 ≤ m ≤ 1000000. The number n…
Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the other knights are fun things to do. Therefore, it is not very surprising that in recent years the kingdom of King Arthur has e…
http://poj.org/problem?id=2942 各种逗.... 翻译白书上有:看了白书和网上的标程,学习了..orz. 双连通分量就是先找出割点,然后用个栈在找出割点前维护子树,最后如果这个是割点那么子树就都是双连通分量,然后本题求的是奇圈,那么就进行黑白染色,判断是否为奇圈即可.将不是奇圈的所有双连通分量的点累计起来即可. #include <cstdio> #include <cstring> #include <cmath> #include <…
题目来源:POJ 2942 Knights of the Round Table 题意:统计多个个骑士不能參加随意一场会议 每场会议必须至少三个人 排成一个圈 而且相邻的人不能有矛盾 题目给出若干个条件表示2个人直接有矛盾 思路:求补图  能够坐在一起 就是能够相邻的人建一条边 然后假设在一个奇圈上的都是满足的 那些不再不论什么一个奇圈的就是不满足 求出全部奇圈上的点 总数减去它就是答案 首先有2个定理 1.一个点双连通分量是二分图 你就没有奇圈 假设有奇圈  那就不是二分图  充分必要条件 所…