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题目的意思就是在1到n的所有序列之间,找出所有相邻的数相加是素数的序列。Ps:题目是环,所以头和尾也要算哦~
典型的dfs,然后剪枝。
这题目有意思的就是用java跑回在tle的边缘,第一次提交就tle了(服务器负载的问题吧),一模一样的第二次提交就ac了,侧面也反应了递归对stack的开销影响效率也是严重的。好了,上代码!!
题目传送门:
import java.util.Scanner; public class Main { public static final int[] prime = { 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1 }; public static boolean[] visited; public static int[] nums; public static int n; public static void dfs(int deepth,int current){ if(deepth == n ){ if(prime[current+1]==1){ for(int i=0;i<n;i++){ System.out.print( nums[i]); if(i+1!=n){ System.out.print( " " ); } } System.out.println( ); } } else { for(int i=2;i<=n;i++){ if(!visited[i] && prime[current+i]==1){ visited[i]=true; nums[deepth] = i; dfs(deepth+1,i); visited[i]=false; } } } } public static void main( String[] args ) { Scanner sc = new Scanner( System.in ); int c=1; while( sc.hasNext() ) { n = sc.nextInt(); nums = new int[ n+1 ]; visited = new boolean[ n+1 ]; nums[0]=1; System.out.println("Case "+c+":"); dfs(1,1); c++; System.out.println( ); } } }
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原文地址:http://www.cnblogs.com/dick159/p/5110936.html