Catenyms】的更多相关文章

Catenyms Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8756   Accepted: 2306 Description A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For e…
Catenyms Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8173   Accepted: 2149 Description A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For e…
UVA 10441 - Catenyms 题目链接 题意:给定一些单词,求拼接起来,字典序最小的,注意这里的字典序为一个个单词比过去,并非一个个字母 思路:欧拉回路.利用并查集判联通,然后欧拉道路判定,最后dfs输出路径 代码: #include <cstdio> #include <cstring> #include <string> #include <vector> #include <iostream> #include <algo…
                                                           Catenyms Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11526   Accepted: 2993 Description A catenym is a pair of words separated by a period such that the last letter of the fi…
// 很久不写图论,连模板也不熟悉了0.0 // 这题是一个技巧性比较高的暴力DFS Catenyms 题目大意 定义catenym为首尾字母相同的单词组成的单词对, 例如: dog.gopher gopher.rat rat.tiger aloha.aloha arachnid.dog 而catenym系统则是多个利用该规则组成的单词串,例如: aloha.aloha.arachnid.dog.gopher.rat.tiger 给定一组单词,让你判断是否能组成catenym系统,且每个单词只能…
A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For example, the following are catenyms: dog.gopher gopher.rat rat.tiger aloha.aloha arachnid.dog A compound…
Description A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For example, the following are catenyms: dog.gopher gopher.rat rat.tiger aloha.aloha arachnid.dog…
题意:给n个字符串(3<=n<=1000),当字符串str[i]的尾字符与str[j]的首字符一样时,可用dot连接.判断用所有字符串一次且仅一次,连接成一串.若可以,输出答案的最小字典序(dot是最小字典序的,比‘a'小). 显然就是以26个字母为结点,n个字符串为边,求解有向图的欧拉通路. 不过这里要注意,26个字母不一定都用上. 先判断有向图的欧拉通路的条件是否成立: 1.有一个结点入度等于出度+1且有一个结点出度等于入度+1且其他结点入度等于出度.(或所有结点入度等于出度) 2.有向图…
poj2337:http://poj.org/problem?id=2337 题意:给定一些单词,如果一个单词的尾字母与另一个的首字母相同则可以连接.问是否可以每个单词用一次,将所有单词连接,可以则输出字典序最小的序列. 题解:并查集+欧拉通路+贪心思维+dfs ,这一题我也是参考了别人的代码.   ps:vector的使用 ,内部堆栈的使用 #include<iostream> #include<cstring> #include<algorithm> #includ…
题意: 就是给出几个单词 看能否组成欧拉回路或路径  当然还是让输出组成的最小字典序的路 解析: 还是把首尾字母看成点   把单词看成边 记录边就好了 这题让我对fleury输出最小字典序又加深了一些认识 fleury输出最小字典序  就必须保证对应输出的边或点  按从小到大的顺序去走 所以我们先保存  然后排序  然后从大到小加边 因为我们用的是邻接表  邻接表是从当前起点u的最后一个加入的边 开始的 ..所以我们要对应起来 把边从大到小依次加入 #include <iostream> #i…