UVa 524 Prime Ring Problem (回溯)

第一次无指导写的回溯。

感觉也不难,小紫书上说“学习回溯短则数天,长则数月或一年以上”, 但我没用一小时就懂了回溯,不知道是真懂还是假懂。

这道题很简单。细心就好。

代码如下:

  1. #include <iostream>
  2. #include <cstdio>
  3. #include <cstring>
  4. #include <cmath>
  5. using namespace std;
  6. int n, ans[25];
  7. const int pn[] = {1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41};
  8. bool isPrimeNumber(int x) {
  9. for (int i = 0; i != 14; ++i) {
  10. if (x == pn[i]) return true;
  11. }
  12. return false;
  13. }
  14. void dfs(int cur) {
  15. if (cur + 1 == n) {
  16. for (int i = 0; i != n - 1; ++i) {
  17. printf("%d ", ans[i]);
  18. }
  19. printf("%d\n", ans[n - 1]);//结尾不能多空格
  20. } else {
  21. for (int i = 1; i != n + 1; ++i) {
  22. bool ok = true;
  23. for (int j = 0; j != cur + 1; ++j) {
  24. if (i == ans[j]) {
  25. ok = false;
  26. break;
  27. }
  28. }
  29. if (!ok) continue;
  30. if ((cur != n - 2 || (cur == n - 2 && isPrimeNumber(i + 1)) )
  31. && isPrimeNumber(i + ans[cur])) {
  32. ans[cur + 1] = i;
  33. dfs(cur + 1);
  34. }
  35. }
  36. }
  37. }
  38. int main() {
  39. int kase(0);
  40. while (scanf("%d", &n) == 1) {
  41. if (kase) printf("\n");//最后一行不能多空行,否则WA
  42. memset(ans, 0, sizeof(ans));
  43. printf("Case %d:\n", ++kase);
  44. ans[0] = 1;
  45. dfs(0);
  46. }
  47. return 0;
  48. }

来自为知笔记(Wiz)

时间: 2024-08-16 18:34:19

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