HDU 1016 Prime Ring Problem(DFS)

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)
Total Submission(s): 18110    Accepted Submission(s): 8109

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)

 

Recommend

JGShining

 

 

一开始想到用BFS搜索，但是标志设定是个大问题，想了很久还是没想出解决方法，经同学指点之后，我忽然明白，DFS可以解决这个问题，果断换上DFS，WA了几次，对比发现那case没大写。。。。囧

 

贴下龌龊的代码：

 

#include<stdio.h>
#include<string.h>
#include<math.h>
int vis[25], pr[104];
int n;
void prime()
{
    int m = 10, i, j;
    memset(pr, 0, sizeof(pr));
    for(i = 2; i <= m; i++)
    {
        if(!pr[i])
        {
            for(j = i * i; j <= 100; j += i)
                pr[j] = 1;
        }
    }
}
void dfs(int g[], int cur)
{
    int i, j;
    for(i = 2; i <= n; i++)
    {
        if(vis[i] || pr[g[cur - 1] + i])continue;
        g[cur] = i;
        if(cur + 1 == n)
        {
            if(pr[g[cur] + 1])continue;
            j = 0;
            printf("%d", g[j]);
            for(j = 1; j <= cur; j++)
                printf(" %d", g[j]);
            printf("\n");
            return ;
        }
        vis[i] = 1;
        dfs(g, cur + 1);
        vis[i] = 0;
    }
}
int main()
{
    int count = 0, g[30];
    prime();
    while(scanf("%d", &n) != EOF)
    {
        printf("Case %d:\n", ++count);
        memset(g, 0, sizeof(g));
        memset(vis, 0, sizeof(vis));
        g[0] = 1;
        dfs(g, 1);
        printf("\n");
    }
    return 0;
}