• HDU 1016 Prime Ring Problem (回溯法)


    Prime Ring Problem

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 34846    Accepted Submission(s): 15441


    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
    题意:在1到n构成一个圆环两两相邻的数和为素数,输出所有情况。
    思路:感觉搜索最重要的就是找状态,这里的状态就是(当前这个数,已经放置了的个数)。
    收获:回溯法:因为有可能再次使用一个数,所以回溯。
    #include <cstdio>
    #include <iostream>
    #include <cstdlib>
    #include <algorithm>
    #include <ctime>
    #include <cmath>
    #include <string>
    #include <cstring>
    #include <stack>
    #include <queue>
    #include <list>
    #include <vector>
    #include <map>
    #include <set>
    using namespace std;
    
    const int INF=0x3f3f3f3f;
    const double eps=1e-10;
    const double PI=acos(-1.0);
    #define maxn 500
    
    int n;
    int vis[maxn];
    int a[maxn];
    int judge(int x)
    {
        if(x <= 1) return 0;
        int m = floor(sqrt(x) + 0.5);
        for(int i = 2; i <= m; i++)
            if(x%i == 0) return 0;
        return 1;
    }
    void dfs(int pos, int num)
    {
        a[num] = pos;
        if(num == n && judge(pos+1))
        {
            for(int j = 1; j <= n-1; j++)
                printf("%d ", a[j]);
            printf("%d
    ", a[n]);
            return;
        }
        for(int i = 2; i <= n; i++)
        {
            int sum  = pos + i;
            if(judge(sum) && !vis[i])
            {
                vis[i] = 1;
                dfs(i, num+1);
                vis[i] = 0;
            }
        }
    }
    int main()
    {
        int cas = 1;
        while(~scanf("%d", &n))
        {
            flag = 1;
            memset(vis, 0, sizeof vis);
            printf("Case %d:
    ", cas++);
            dfs(1, 1);
            puts("");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ZP-Better/p/4639614.html
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