HDU 1016 Prime Ring Problem(搜索入门题 DFS,多种方式)

Prime Ring Problem Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 39066    Accepted Submission(s): 17239 Problem Description 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 always be 1.   Input n (0 < n < 20).   Output The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order. You are to write a program that completes above process. Print a blank line after each case.   Sample Input 6 8   Sample Output Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2   Source Asia 1996, Shanghai (Mainland China) 题意: n个数(1-n)构成一个素数环,是的相邻两个数的和均为素数,第一个数固定为1. 思路:简单深搜.下面三份代码,一二份搜索的参数都只有一个,而最后一份有两个, 搜索的参数由自己定,但考虑入门的人(像当初的我),为他们提供参考. AC代码: #include<iostream> #include <cstring> using namespace std; int n; int a[25]; bool vis[25]; bool prime(int x) { for(int i=2;i*i<=x;i++) if(x%i==0) return false; return true; } void DFS(int x) { if(x==n&&prime(a[x-1]+a[0])) //搜索(递归)结束条件:找到了最后一个,并且最后一个与第一个(a[0],也就是1)的和也为素数 { for(int i=0;i<n;i++) { if(i) cout<<" "; cout<<a[i]; } cout<<endl; return ; } for(int i=1;i<=n;i++)//1到n的数依次去搜索 { if(!vis[i]&&prime(a[x-1]+i))//i没用过且与上一个数的和为素数 { vis[i]=true;//标记 i 用过 a[x]=i;//第x个数为i DFS(x+1);//递归,搜索第x+1个 vis[i]=false;//取消标记.使得下次搜索能进行,当前搜索(n个数组合的一种情况)结束时(无论成功与否) } } } int main() { int kase=0; while(cin>>n) { cout<<"Case "<<++kase<<":"<<endl; a[0]=1;//第0个数为 1 memset(vis,false,sizeof(vis));//vis[i]=false-->i未用过 vis[1]=true;//标记 1 用过 DFS(1);//搜索第1个数 cout<<endl; } return 0; } 上面那份代码的数组元素下标从0开始,下面这份代码的从1开始, #include <stdio.h> #include <string.h> int n; int a[25]; int visit[25]; int sushu(int x); int DFS(int x); int main() { int cas=0; a[1]=1; while (scanf("%d",&n)!=EOF) { printf("Case %d:\n",++cas); memset(visit,0,sizeof(visit)); visit[1]=1; DFS(2); printf("\n"); } return 0; } int sushu(int x) { int i; for (i=2;i*i<x+1;i++) { if (x%i==0) return 0; } return 1; } int DFS(int x) { int i; if (x==n+1&&sushu(a[n]+a[1])==1) { for (i=1;i<n;i++) { printf("%d ",a[i]); } printf("%d\n",a[n]); return ; } for (i=2;i<=n;i++) { if (visit[i]==0) { if (sushu(a[x-1]+i)==1) { visit[i]=1; a[x]=i; DFS(x+1); visit[i]=0; } } } } 两个参数: #include <stdio.h> #include <string.h> #include <stdbool.h> int n; bool vis[25];//标记各个数是否用过 int num[25];//填的数字 bool fun(int s) { for(int i=2; i*i<=s; i++) { if(s%i==0) return false; } return true; } void Dfs(int s,int t)//s第t-1个数的值 t要搜索第t个数 { if(t==n+1)//跳出的条件 { if(fun(s+1)==true) { for(int i=1; i<n; i++) { printf("%d ",num[i]); } printf("%d\n",num[n]); } return ; } for(int i=1; i<=n; i++)//1到n中的数去寻找 { if(vis[i]==false)//判断是否重复出现//即 没有用过 { if(fun(s+i)==true) { num[t]=i; vis[i]=true; Dfs(i,t+1); vis[i]=false;////进入到最深层后，返回时用到 } } } } int main() { int Case=0; while(~scanf("%d",&n)) { printf("Case %d:\n",++Case); memset(vis,0,sizeof(vis)); vis[1]=true;//确定这个数字使用过 num[1]=1;//输出的数字 Dfs(1,2); printf("\n"); } return 0; }