uva 696 - How Many Knights

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uva 696 - How Many Knights

题目链接:uva 696 - How Many Knights 题目大意:给出一个n * m的网格,计算最多可以放置几个国际象棋中的骑士. 解题思路:分成三类来讨论: 1)min(n, m) == 1, 也就是无论怎么摆也不会影响到其他的骑士. 2)min(n, m) == 2, 这是将网格将网格分成2*4的若干部分,每个部分的前半部分放置骑士,主要注意模4后剩余部分的处理. 3)n *m的网格上间隔摆放(就是对应的黑格子或者白格子)(n * m - 1)/ 2 #include <stdio.…

UVA题目分类

题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics 10300 - Ecological Premium 458 - The Decoder 494 - Kindergarten Counting Game 414 - Machined Surfaces 490 - Rotating Sentences 445 - Marvelous Mazes…

uva 3523 Knights of the Round Table

题意:给你n,m n为有多少人,m为有多少组关系,每组关系代表两人相互憎恨,问有多少个骑士不能参加任何一个会议. 白书算法指南对于每个双联通分量,若不是二分图,就把里面的节点标记 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <…

UVA 1364 - Knights of the Round Table （获得双连接组件 + 二部图推理染色）

尤其是不要谈了些什么,我想A这个问题! FML啊.....! 题意来自 kuangbin: 亚瑟王要在圆桌上召开骑士会议.为了不引发骑士之间的冲突. 而且可以让会议的议题有令人惬意的结果,每次开会前都必须对出席会议的骑士有例如以下要求: 1. 相互憎恨的两个骑士不能坐在直接相邻的2个位置: 2. 出席会议的骑士数必须是奇数,这是为了让投票表决议题时都能有结果. 注意:1.所给出的憎恨关系一定是双向的.不存在单向憎恨关系. 2.因为是圆桌会议.则每一个出席的骑士身边必然刚好有2个骑士. 即…

UVA它11292 - Dragon of Loowater

Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predato…

UVA 11292 Dragon of Loowater(简单贪心)

Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predato…