hdu 6311 欧拉回路

【hdu 6311 欧拉回路】的更多相关文章

题意:求一个图(不一定联通)最小额外连接几条边,使得可以一笔画出来大致做法 1.找出联通块 2.统计每一个连通块里面度数为奇数的点的个数, 有一个性质一个图能够用一笔画出来,奇数点的个数不超过2个 if 奇数点的个数==0 或者 ==1 直接找欧拉回路 else 将除去前面两个奇数点外的奇数点依次相连然后找欧拉回路然后记录路径 #include <cstdio> #include <cstring> #include <queue> #include <…

HDU 6311 Cover （无向图最小路径覆盖）

HDU 6311 Cover (无向图最小路径覆盖) Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1967 Accepted Submission(s): 442 Special Judge Problem Description The Wall has down and the King in the north has to sen…

HDU - 6311:Cover（欧拉回路，最少的一笔画覆盖无向图）

The Wall has down and the King in the north has to send his soldiers to sentinel. The North can be regard as a undirected graph (not necessary to be connected), one soldier can cover one path. Today there's no so many people still breathing in the no…

hdu 1116 欧拉回路+并查集

http://acm.hdu.edu.cn/showproblem.php?pid=1116 给你一些英文单词,判断所有单词能不能连成一串,类似成语接龙的意思.但是如果有多个重复的单词时,也必须满足这样的条件才能算YES.否则都是不可能的情况. 欧拉回路和欧拉通路的判定可以总结为如下: 1)所有的点联通 2)欧拉回路中所有点的入度和出度一样. 3)欧拉通路中起点的入度 - 出度 = 1,终点的初度 - 入度 = 1, 其他的所有点入度 = 出度: 所以用并查集搞就好了 #pragma comm…

HDU 1878 欧拉回路（判断欧拉回路）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1878 题目大意:欧拉回路是指不令笔离开纸面,可画过图中每条边仅一次,且可以回到起点的一条回路.现给定一个图,问是否存在欧拉回路? 解题思路:判断无向图是否存在欧拉回路,判断每个点的度数是否为偶数+并查集确认连通性. 代码: #include<iostream> #include<cstring> #include<cstdio> #define CLR(arr,val) m…

HDU 3018 欧拉回路

HDU - 3018 Ant Country consist of N towns.There are M roads connecting the towns. Ant Tony,together with his friends,wants to go through every part of the country. They intend to visit every road , and every road must be visited for exact one time.Ho…