回溯法确实不是很好理解掌握的,学习紫书的代码细细体会. #include <cstdio> ]; int n, L, cnt; int dfs(int cur) { if(cnt++ == n) { ; i < cur; ++i) { == && i) puts(""); == && i) printf(" "); printf("%c", 'A' + S[i]); } printf("…

原题:https://uva.onlinejudge.org/external/1/129.pdf 按照字典顺序生成第n个“困难的串” “困难的串”指的是形如ABAB, ABCABC, CDFGZEFGZE的串,它们都有相邻的重复子字符串 字母的范围是L,既 'A'到'A' + L 分析: 大体上这是一道生成排列组合的题.难点在于我们如何判断当前生成的串是"困难的串" 我们首先采用递归按照字典顺序从小到大生成串, 那么每一次我们处理的都是前一个"困难的串", 既满足…

UVA.129 Krypton Factor (搜索+暴力) 题意分析 搜索的策略是:优先找长串,若长串不合法,则回溯,继续找到合法串,直到找到所求合法串的编号,输出即可. 注意的地方就是合法串的判断,根据后缀的规则来判断,枚举后缀长度[1,len/2],后缀中是否有重复子串,若是的话表明不是合法串. 还有一个注意的地方,每次递归调用时,序号就要+1,无论是回溯回来的递归,还是深度搜索的递归,因为没找到一组可行解,编号就要加一.原因是此题不是用递归深度来判断输出的. 代码总览 #include…

回溯法,只需要判断当前串的后缀,而不是所有的子串 #include<iostream> #include<cstdio> using namespace std; ]; int n,l,cnt; int bfs(int cur) { if(cnt++==n) { ; i<cur; i++) printf("%c",'A'+s[i]); cout<<endl; ; } ; i<l; i++) { s[cur]=i; ; ; j*<=c…

 Krypton Factor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 392    Accepted Submission(s): 174 Problem Description You have been employed by the organisers of a Super Krypton Factor Conte…

 Krypton Factor  You have been employed by the organisers of a Super Krypton Factor Contest in which contestants have very high mental and physical abilities. In one section of the contest the contestants are tested on their ability to recall a seque…

UVA - 524 Prime Ring Problem Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  into each circle separately, and the sum of number…

/*UVa129 - Krypton Factor --回溯问题.看例子可知道确定该字符串是按照从左到右依次考虑每个位置,当前位置填不上所有的字符时,需要回溯. -- */ #define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<fstream> #include<time.h> #include<vector> using namespace std; const int maxn = 10…

题意:输入n,把1~n组成个环,相邻两个数之和为素数. 分析:回溯法. #pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<s…

[10]Regular Expression Matching [17]Letter Combinations of a Phone Number [22]Generate Parentheses (2019年2月13日) 给了一个N,生成N对括号的所有情况的字符串. n = 3 [ "((()))", "(()())", "(())()", "()(())", "()()()" ] 题解:dfs生成. c…