Chinese Postman Problem is a very famous hard problem in graph theory. The problem is to find a shortest closed path or circuit that visits every edge of a (connected) undirected graph. When the graph has an Eulerian Circuit (a closed walk that cover…

Prime Ring Problem                                                                             Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)                                                                        …

The Chinese Postman Problem My Tags   (Edit)   Source : bin3   Time limit : 1 sec   Memory limit : 64 M Submitted : 503, Accepted : 172 A Chinese postman is assigned to a small town in China to deliver letters. In this town, each street is oriented a…

本题题意:寻找一个排列,在此排序中,带宽的长度最小(带宽是指:任意一点v与其距离最远的且与v有边相连的顶点与v的距离的最大值),若有多个,按照字典序输出最小的哪一个. 解题思路: 方法一:由于题目说结点的个数最多是8个,所以,最先想到的方法是暴力枚举,将所有的结点全排列,然后找到宽带长度最小的那一个,此方法不会超时. 方法二:利用回溯法,将所有肯能的情况都遍历一遍,保存目前最短的宽带的长度minn,若当前的宽带长度大于minn,就进行剪枝,此题要注意,此题的结点不一定是从A开始的,所以要建立字母…

呜呜呜 递归好不想写qwq 求“所有情况”这种就递归 17. Letter Combinations of a Phone Number 题意:在九宫格上按数字,输出所有可能的字母组合 Input: " Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf&qu…

40. 组合总和 II LeetCode_40 题目描述 题解分析 此题和 39. 组合总和 + 递归 + 回溯 + 存储路径很像,只不过题目修改了一下. 题解的关键是首先将候选数组进行排序,然后记录每个数的出现次数. 将去重后的数组当成是新的候选数组进行递归搜索. 回溯的时候注意是在最后将相同数字次数的数从列表中清除. java代码 package com.walegarrett.interview; import java.util.ArrayList; import java.util.A…

题目链接:http://acm.hit.edu.cn/hoj/problem/view?id=2739 Time limit : 1 sec Memory limit : 64 M A Chinese postman is assigned to a small town in China to deliver letters. In this town, each street is oriented and connects exactly two junctions. The postma…

题目链接:http://acm.fzu.edu.cn/problem.php?pid=2224 同hdu5869 //#pragma comment(linker, "/STACK:102400000, 102400000") #include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #inc…

博弈,$SG$函数,规律,线段树. 这个问题套路很明显,先找求出$SG$函数值是多少,然后异或起来,如果是$0$就后手赢,否则先手赢.修改操作和区间查询的话可以用线段树维护一下区间异或和. 数据那么大,一看就知道$SG$有规律...... 先写个小数据的$SG$找规律: ]; ]; int SG(int x) { memset(f,,sizeof f); ;i>=x/;i--) { if(x-i>i) break; f[sg[i]]=; } ;i<=;i++) { ) continue;…

 Bracket Sequences Concatenation Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A bracket sequence is a string containing only characters "(" and ")". A regular bra…