1）注意格式要求(@1234)；先順時針輸出再逆時針輸出並不用儲存所有可能再倒著輸出一遍，因為逆時針的輸出被包含在前者之中，這裡我應該是沒讀懂題意；之前在想怎麼執行完一次之後或者找不到之後如何退回到上一步，是不是要return什麼啊之類的，其實執行完遞迴就自動退回了，如果符合邊界條件就輸出，不符合就繼續執行，然後自動退回到上一層，在上一層這個遞迴語句之後只是需要將相應的“標記陣列”之類的，或者cur當前所指，減一即可，回溯的真諦、dfs的真諦，還是理解不透徹；另外需要遞迴時，確定遞迴函式引數，然後在遞迴函式裡寫兩個分支，即if{};（//邊界條件）else{};即可。

#include <iostream>
#include <math.h>
#include <string.h>

using namespace std;
int n;int g;
int kase;
int bj[23]={0};
int jg[23];
//int jilu[23][23];
int pri[41];//={1},youxiao ma
void dfs(int cur)
{
    if(cur==n&&pri[jg[n-1]+jg[0]]){
        for(int i=0;i<n-1;i++){
            cout<<jg[i]<<" ";//@2
        }
        cout<<jg[n-1]<<endl;//@3
        //memcpy(&jg,&jilu[g++],sizeof(jg));
    }
    else{
        for(int i=2;i<=n;i++){
            if(!bj[i]&&pri[jg[cur-1]+i]){
                bj[i]=1;
                jg[cur]=i;
                dfs(cur+1);//不用cur++，則從這遞迴完成出來以後也不用cur--；
                bj[i]=0;
            }
        }
    }
}
bool is_pri(int k){
    for(int i=2;i<=sqrt(k);i++){
        if(k%i==0){
            return 0;
        }
    }
    return 1;
}
int main()
{
    //素數提前打表，提高速度
    for(int i=2;i<=40;i++){
        if(is_pri(i))
            pri[i]=1;
    }
    kase=0;
    while(cin>>n){
        printf("Case %d:\n",++kase);//@1
        jg[0]=1;
        dfs(1);
        cout<<endl;//@4
        //prin();
    }
}
 

2）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38645    Accepted Submission(s): 17074


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