Day2-M-Prime Ring Problem-HDU1016

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. 

 

Inputn (0 < n < 20). 
OutputThe 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

思路：依旧是回溯法，逐个判断即可，注意首尾也需要判断一下，用筛法建个表，代码如下：

const int maxm = 22;

int n, vis[maxm], prime[330], res[maxm], kase = 0;

void buildTable() {
    for (int i = 2; i * i < 330; ++i) {
        if(!prime[i]) {
            for (int j = i * i; j < 330; j += i)
                prime[j] = 1;
        }
    }
}

void dfs(int i) {
    if(i == n && !prime[res[0] + res[n-1]]) {
        for (int j = 0; j < n; ++j) {
            if(j)printf(" ");
            printf("%d", res[j]);
        }
        printf("\n");
    }
    for (int j = 2; j <= n; ++j) {
        if(!vis[j] && !prime[res[i-1] + j]) {
            vis[j] = 1;
            res[i] = j;
            dfs(i + 1);
            vis[j] = 0;
        }
    }
}


int main() {
    buildTable();
    while(scanf("%d",&n) != EOF) {
        printf("Case %d:\n", ++kase);
        memset(vis, 0, sizeof(vis));
        res[0] = 1;
        vis[1] = 1;
        dfs(1);
        printf("\n");
    }
    return 0;
}

View Code

 

标签：Prime,prime,int,res,330,maxm,HDU1016,Ring,circle
来源： https://www.cnblogs.com/GRedComeT/p/11231268.html