(Hdu step 4.3.2) Prime Ring Problem (n number is a Ring, and all the conditions for the sum of the two is output), hdu4.3.2

(Hdu step 4.3.2) Prime Ring Problem (n number is a Ring, and all the conditions for the sum of the two is output), hdu4.3.2

Question:

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission (s): 467 Accepted Submission (s): 297

Problem DescriptionA ring is compose of n circles as shown in digoal. put natural number 1, 2 ,..., n into each circle separately, and the sum of numbers in two adjacent circles shoshould be a prime. Note: the number of first circle shoshould always be 1.

Inputn (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 abve process. Print a blank line after each case.

Sample Input 68

Sample Output Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

SourceAsia 1996, Shanghai (Mainland China)

RecommendJGShining

Question Analysis:

Simple question. Use DFS. Note that the nth number also needs to determine whether the sum of sum 1 is a prime number. It should be noted that, compared with the previous question, this question is a one-dimensional DFS question, while the last question is a two-dimensional DFS question. Therefore, this question is not applied to map [] [] of the Bao Cun map information.

DFS rules are as follows:

1) access an adjacent unaccessed node and import it to the stack

2) If rule 1 cannot be executed, an element of the stack is output.

3) If rule 1 cannot be executed, 2) the traversal is completed.

The Code is as follows:

/** B. cpp ** Created on: February 24, 2015 * Author: Administrator */# include <iostream> # include <cstdio> # include <algorithm> # include <cmath> using namespace std; const int maxn = 21; int value [maxn]; // used to store the final conforming result queue bool visited [maxn]; // indicates whether a certain number has accessed int n./*** determines whether a certain number n is a prime number */bool isPrime (int n) {int I; for (I = 2; I <= sqrt (n) + 0.5; ++ I) {// If a number is a combination, name must have a factor that is <sqrt (n) if (n % I = 0) {// if it can remove ireturn fal Se; // It indicates that it is not a prime number} return true; // It indicates that it is a prime number}/*** output result queue meeting the requirements */void printValues () {int I; for (I = 1; I <n; ++ I) {printf ("% d", value [I]);} printf ("% d \ n ", value [n]);}/*** Deep Search * cnt: number of currently searched steps */void dfs (int cnt) {// if n steps have been completed, and the sum of the number of cut n and 1st is the prime number if (cnt = (n + 1) & isPrime (value [cnt-1] + 1) {printValues (); // name: this is the case that meets the requirements. Output all results} int I; for (I = 2; I <= n; ++ I) {// traverse each number, check whether it is suitable to be added to the current result queue. // If I has not been accessed & the last number and I of the current result Sequence And is a prime number if (visited [I] = false) & isPrime (I + value [cnt-1]) {visited [I] = true; // mark I as value [cnt] = I; // Add I to the result sequence dfs (cnt + 1 ); // search for the next suitable number visited [I] = false along the sequence; // the above case has been taken into consideration and I is reset to inaccessible. used to consider the next situation }}int main () {int counter = 1; while (scanf ("% d", & n )! = EOF) {memset (visited, false, sizeof (visited); printf ("Case % d: \ n", counter); value [1] = 1; // The 1st Number of sequences is always 1 visited [1] = true; dfs (2); // search printf ("\ n") from step 1 "); counter ++;} return 0 ;}