HDU1016:Prime Ring Problem(DFS)

dfs培养的是一种思维，
简单点来说用dfs做最短路一般会超时，所以更多会要是思维的训练

Prime Ring Problem
Time Limit: 2000msMemory Limit: 32768KB This problem will be judged on HDU. Original ID: 1016
64-bit integer IO format: %I64d Java class name: Main
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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
这题说形成素数环，其实可看为，后一位的前两位相加为素数，一直查找到n+1位；

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int prime[100] = {0,0,1};int n;int b[100];int stu[100];int k = 0;void dfs(int depth){    if (depth == n + 1 && prime[b[1] + b[n]])    {        for (int i = 1; i <= n; i++)        {            if (i == 1)                cout << b[i];            else                cout << " " << b[i];        }        puts("");        return;    }    for (int i = 2; i <= n; i++)    {        if (!stu[i] && prime[i + b[depth - 1]])        {            stu[i] = 1;            b[depth] = i;            dfs(depth + 1);            stu[i] = 0;        }    }}int main(){     //prime    int i, j;    for (i = 3; i < 100; i++)    {        prime[i++] = 1;        prime[i] = 0;    }    for (i = 3; i < 100; i++)        for (j = i + i; j < 100; j += i)            prime[j] = 0;    //int k = 0;    while (cin>>n)    {        printf("Case %d:\n", ++k);    //  puts("");        memset(stu, 0, sizeof(stu));        b[1] = 1;        stu[1] = 1;        dfs(2);        puts("");    }    return 0;}