one recursive approach to solve hdu 1016, list all permutations, solve N-Queens puzzle. reference: the video of stanford cs106b lecture 10 by Julie Zelenski https://www.youtube.com/watch?v=NdF1QDTRkck // hdu 1016, 795MS #include <cstdio> #include &l…

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42600    Accepted Submission(s): 18885 Problem Description A ring is compose of n circles as shown in diagram. Put natural num…

HDOJ(HDU).1016 Prime Ring Problem (DFS) [从零开始DFS(3)] 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DFS HDOJ.1010 Tempter of the Bone [从零开始DFS(1)] -DFS四向搜索/奇偶剪枝 HDOJ(HDU).1015 Safecracker [从零开始DFS(2)] -DFS四向搜索变种 HDOJ(HDU).1016 Prime Ring Problem (…

Prime Ring Problem HDU - 1016 A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime. Note: the number of first circle should a…

http://acm.hdu.edu.cn/showproblem.php?pid=1016 #include <iostream> using namespace std; int a[30],n,used[40]; int is_prime(int x) { for(int i=2;i<x; i++) if(x%i==0) return 0; return 1; } void dfs(int cur) { int i; if(cur==n && is_prime(a[…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1016 Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description A ring is compose of n circles as shown in diagram. Put natural number 1, 2…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1016 Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 63806    Accepted Submission(s): 27457 Problem Description A ring is compos…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1016 题目大意:输入一个n,环从一开始到n,相邻两个数相加为素数. #include <iostream> #include <cstdio> #include <cstring> using namespace std; ],q[],n; int sushu(int n) { int i; ; i*i<=n; i++) ) ; ; } void dfs(int x,…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1016 哈哈,状态来了就是不一样,很快就A了. 注意的位置是,最后一个点,要与第一个点比较. #include <stdio.h> #include <string.h> ]= {}; ///是素数就是0 ]; ]; int n; void dfs(int k) { //printf("%d\n",k); ) { ; i<n; i++) printf("…

嗯... 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1016 一道很典型的dfs+回溯: 根据题意首先进行初始化,即第一个位置为1,然后进行dfs,枚举2~n之间的每一个数,如果这个数没被使用并且它和环中上一个数形成素数环,那么就把它加入环中,打上标记,然后继续dfs,最后回溯.当环上的个数正好等于n并且第一个数和最后一个数也能组成素数,则输出,输出时注意格式,很严格! dfs这里还有一个剪枝: 只有n为偶数时才可能形成素数环!因为当n为奇数时…