hdu 1016 Problem Description (广搜)

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
 
 
还是一道广搜的题目，貌似还是花了很长的时间。
虽然说自己的做法可能有些啰嗦，没有大神们的代码精简。
不过还是贴出来，留着以后自己再慢慢优化吧。
View Code 
 1 #include<iostream>
 2 #include<math.h>
 3 using namespace std;
 4 
 5 int num[30];
 6 int N , Count;
 7 int vis[30];
 8 int sushu[100];
 9 int fun(int n)
10 {
11     int i , flag1 = 1 , flag2 = 1;
12     for(i = 2; i <=n; i++)
13     {
14         int temp = num[i] + num[i-1];
15         if(!sushu[temp])
16             flag1 = 0;
17     }
18     if(n == N)
19     {
20         int temp = num[N] + 1;
21         if(!sushu[temp])
22             flag2 = 0;
23     }
24     if(flag1 && flag2) return 1;
25     else return 0;
26 }
27 
28 int dfs(int *num , int n)
29 {
30     int i , j;
31     for(i = 2; i <= N; i++)
32     {
33         if(vis[i]) continue;
34         num[n] = i;
35         vis[i] = 1;
36         if(n>2 && !fun(n)) 
37         {
38             vis[i] = 0;
39             continue;
40         }
41         if(n == N && fun(n))
42         {
43             for(j = 1; j <= N; j++)
44             {
45                 cout<<num[j];
46                 if(j != N) cout<<' ';
47             }
48             cout<<endl;
49             vis[i] = 0;
50             continue;
51         }
52         dfs(num , n+1);
53         vis[i] = 0;
54     }
55     return 0;
56 }
57 
58 
59 void find()  //生成素数
60 {
61     int i;
62     sushu[1] = 1;
63     sushu[2] = 1;
64     for(i = 3; i <100; i++)
65     {
66         int temp , j , flag = 0;
67         temp = (int)sqrt(double(i));
68         for(j = 2; j <= temp; j++)
69         {
70             if(i % j == 0)
71                 flag = 1;
72         }
73         if(!flag) sushu[i] = 1;
74     }
75 }
76 int main()
77 {
78     num[1] = 1;
79     memset(sushu , 0 , sizeof(sushu));
80     find();
81     Count = 1;
82     while(cin>>N)
83     {
84         cout<<"Case "<<Count<<":"<<endl;
85         Count++;
86         memset(vis , 0,sizeof(vis));
87         vis[1] = 1;
88         dfs(num , 2);
89         cout<<endl;
90     }
91     return 0;
92 }

 

posted on 2013-05-06 21:28  nigel_jw  阅读(146)  评论(0)    收藏  举报

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