#HDU 1016 Prime Ring Problem 【DFS+溯回求组数】
2016-02-23 00:37
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题目:
Prime Ring Problem
[b]Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 38251 Accepted Submission(s): 16926
[/b]
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
题意即,给定数字N,将1~N共N个数字排序,使得相邻两数之和为质数,求出所有符合的排序。
首先我们要求出20以内所有质数制成质数表,方便查询。
对于每次搜索,初始值一定为1(题目规定),第二位有(N-1)种选择,即从初始状态可DFS递推至(N-1)条分路。
对于每一条路,又可以有(N-2)种选择。为了减少复杂度,当我们发现待选的路径和上一位不满足加和为质数时,我们将对这条路进行剪枝。即直接break掉,溯回上一层重新选择。
最后输出所有可能即可。
#include<iostream> #include<algorithm> #include<string.h> using namespace std; int data[25][25]; int ans[25]; int que[25], n, be; int sim[20] = { 0, 2, 3, 5, 7, 11, 13, 17, 19 }; int finding(int a, int b) { for (size_t i = 2; i <= 8; i++) { if ((a + b) % i == 0 && (a + b) != i)//建立质数表 { return 1; } } return 0; } void fins() { for (size_t s = 1; s < 21; s++) { for (size_t i = 1; i < 21; i++) { if (!finding(s, i))//按层进行搜索 { data[s][i] = 1; } } } } void print() { for (size_t i = 0; i < n - 1; i++) { cout << que[i] << " ";//que数组用来存已选择的数字,确定满足条件后 按位输出 } cout << que[n - 1]; cout << "\n"; } void dfs(int s) { for (size_t i = 1; i <= n; i++) { if (data[s][i] == 1 && ans[i] == 0) { ans[i] = 1; que[be] = i; be++; if (be == n) { if (data[i][1] == 1) { print(); } } else { dfs(i); } be--; ans[i] = 0; } } } int main() { int cas = 1; fins(); while (cin >> n) { cout << "Case " << cas << ":\n"; be = 0; memset(ans, 0, sizeof(ans)); memset(que, 0, sizeof(que)); ans[1] = 1; que[be] = 1; be++; dfs(1); cout << "\n"; cas++; } }
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