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  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4.  
  5. struct Core{
  6. 	int skills[10];//1 to 9,skills[n]= true if core contains skill n
  7. 	int skillOrder[3];//records the ordering of skills
  8. };
  9. struct SkillOccur {
  10. 	int skill//which skill
  11. 		,occur;//the amount of subjects of this skill
  12. };
  13. void modeOne(struct Core core[], int coreCount, int enumerate,int priority[6]);
  14. void modeTwo(struct Core core[], int coreCount, int enumerate, int priority[6]);
  15. int main() {
  16. 	/*�ѼƳ]�w��*/
  17. 	char mode = '1';/*�]�w'1'��4��6�� ,'2'�� 6��9��*/
  18. 	char enumerate = '2';/*�] '1' �C�X�Ҧ���,�]'2' �C�Ĥ@�զX�G������,�]'3'�C�Ĥ@�ո� */
  19. 	printf("�w��ϥη����������֤߿z�ﾹ~~\n\n");
  20. 	/*�Цb�}�C��1����J�z�������֤ߡA�Ψӧ�|�֤��ޡA�H','���j*/
  21. 	char s1[2000] = "123,143,125,164,136,421,435,415,451,435,431,412,412,432,413,451,652,651,621,615,632,624,623,615,653,523,546,563,562,542,361,352,362,326,356,346,364,324,354,365,264,213,245,236,214";
  22. 	/*�Цb�}�C��2����J�z�������֤ߡA�Ψӧ䤻�֤E�ޡA�H','���j*/
  23. 	char s2[2000] = 
  24. 		 ",629,647,687,361,317,316,368,386,378,386,328,312,394,359,321,394,387,586,549,546,564,583,518,584,538,539,532,518,572,726,768,719,786,764,783,764,746,783,762,243,257,291,476,426,459,431,468,485,459,462,496,473,439,489,429,436,497";
  25. 	char *s;
  26. 	if (mode == '1')s=s1;
  27. 	else s=s2;
  28. 	//967841
  29. 	struct SkillOccur stack[10];
  30. 	int a, b;
  31. 	for (a = 0; a < 10; a++) { stack[a].occur = 0; stack[a].skill = a; }
  32. 	struct Core core[1000];
  33. 	int coreCount = 0;
  34. 	char*p=s;
  35. 	while (1) {
  36.  
  37. 		if (*p == 0)break;
  38. 		if (*p == '0' || *(p + 1) == '0') {
  39. 			stack[*(p + 2) - 48].occur++;
  40. 			p += 4; continue;
  41. 		}
  42. 		else if(*p==',')p++;
  43. 		for (int i = 0; i < 10; i++) {
  44. 			core[coreCount].skills[i] = 0;
  45. 		}
  46. 		for (int i = 0; i < 3; i++) {
  47. 			core[coreCount].skillOrder[i] = *(p)-48;
  48. 				core[coreCount].skills[*(p)-48] = 1;
  49. 				p++;
  50. 		}
  51.  
  52. 		coreCount++;
  53. 	}
  54. 	/*data input done*/
  55.  
  56. 	/*printing cores, recording subject distribution*/
  57.  
  58.  
  59. 	printf("�C�X�֦����֤�...\n");
  60. 	for (a = 0; a < coreCount; a++) {
  61.  
  62. 		printf("�֤߽s��:%d, [ ", a + 1);
  63.  
  64. 		for (b = 0; b < 3; b++) {
  65. 			printf("%d ", core[a].skillOrder[b]);
  66. 			if (b == 2) { stack[core[a].skillOrder[b]].occur += 1; stack[core[a].skillOrder[b]].skill = core[a].skillOrder[b]; }
  67. 		}
  68. 		printf("]\n");
  69. 	}
  70. 	/*bubblesort the stack*/
  71. 	struct SkillOccur temp;
  72. 	for (a = 1; a <= 8; a++) {
  73. 		for (b = 1; b <= 8 - (a-1); b++) {
  74. 			if (stack[b].occur > stack[b + 1].occur) {
  75. 				temp = stack[b+1];
  76. 				stack[b+1] = stack[b];
  77. 				stack[b] = temp;
  78. 			}
  79. 		}
  80. 	}
  81. 	/*�C�X�D���X�{����*/
  82. 	printf("�C�X�D���X�{����\n");
  83. 	b = 9;//the top of the sorted stack
  84. 	for (a = 1; a <= b; a++) printf("�D��%d�X�{%d\n",stack[a].skill,stack[a].occur);
  85. 	int priority[6];
  86. 	if (mode == '1')
  87. 		for (a = 0; a < 4; a++)//we want the top 4 of the stack as priority
  88. 			priority[a] = stack[b - a].skill;
  89. 	else for (a = 0; a <6; a++)//we want the top 6 of the stack as priority
  90. 		priority[a] = stack[b - a].skill;
  91.  
  92. 	/*search*/
  93. 	if (mode == '1')modeOne(core, coreCount,enumerate,  priority);
  94. 	else modeTwo(core, coreCount, enumerate,  priority);
  95.  
  96.  
  97. 	return 0;
  98. }
  99. void modeOne(struct Core core[],int coreCount,int enumerate, int priority[6]) {
  100. 	/*
  101. 	finding core combinations using brute force loops
  102. 	*/
  103. 	int selected[7] = { 0 };
  104. 	int hold[4];
  105. 	int i, j, k, l,a,b;
  106. 	printf("start\n");
  107. 	for (i = 0; i < coreCount; i++) {
  108.  
  109. 		for (a = 1; a <= 6; a++) {
  110. 			if (core[i].skills[a])
  111. 				selected[a] += 1;
  112. 		}
  113. 		hold[0] = i;/*selected 1 core*/
  114. 		for (j = 0; j < coreCount; j++) {
  115. 			if (j == i)continue;
  116. 			for (a = 1; a <= 6; a++) {
  117. 				if (core[j].skills[a])
  118. 					selected[a] += 1;
  119. 			}
  120.  
  121. 			hold[1] = j;//selected 2 cores
  122. 			for (k = 0; k < coreCount; k++) {
  123. 				if (k == i || k == j)continue;
  124. 				int sentinel = 0;
  125. 				for (a = 1; a <= 6; a++) {
  126. 					if (core[k].skills[a] && 3 <= selected[a] + 1)/*if 3== selected[a] + 1, then reject subtree*/
  127. 					{
  128. 						sentinel = 1; break;
  129. 					}
  130.  
  131. 				}
  132. 				if (sentinel)continue;
  133. 				for (a = 1; a <= 6; a++) {
  134. 					if (core[k].skills[a])
  135. 						selected[a] += 1;
  136. 				}
  137.  
  138. 				hold[2] = k;//selected 3 cores
  139. 				for (l = 0; l < coreCount; l++) {
  140. 					if (l == i || l == j || l == k)continue;
  141. 					for (a = 1; a <= 6; a++) {
  142. 						if (core[l].skills[a])
  143. 							selected[a] += 1;
  144. 					}
  145. 					hold[3] = l;//selected 4 cores, checkpoint
  146. 					if (selected[1] == 2//checking condition
  147. 						&& selected[2] == 2
  148. 						&& selected[3] == 2
  149. 						&& selected[4] == 2
  150. 						&& selected[5] == 2
  151. 						&& selected[6] == 2
  152. 						) {
  153. 						//printf("\n找到一組4核六技!\n");
  154. 						int checkDup[4] = { 0,0,0,0 }, dupCount = 0;
  155. 						for (a = 0; a < 4; a++) {
  156. 							//	printf("核心編號:%d, [ ", hold[a] + 1);
  157. 							checkDup[dupCount] = core[hold[a]].skillOrder[2];
  158. 							dupCount++;
  159.  
  160. 							//	printf("]\n\n");
  161.  
  162. 						}
  163. 						if (checkDup[0] != checkDup[1] &&
  164. 							checkDup[0] != checkDup[2] &&
  165. 							checkDup[0] != checkDup[3] &&
  166. 							checkDup[1] != checkDup[2] &&
  167. 							checkDup[1] != checkDup[3] &&
  168. 							checkDup[2] != checkDup[3]
  169. 							) {
  170. 							int prefered = 0;
  171. 							/*compare priority and checkdup*/
  172.  
  173. 							for (a = 0; a < 4; a++)
  174. 								for (b = 0; b < 4; b++)
  175. 									if (priority[a] == checkDup[b])
  176. 										prefered++;
  177. 							if (prefered == 4)
  178. 								printf("\n找到一組四核六技!，且它的主詞符合您持有數量前四多的主詞!(%d,%d,%d,%d)\n", priority[0], priority[1], priority[2], priority[3]);
  179. 							/*else 
  180. 								printf("\n找到一組四核六技!，但它的主詞不符合您持有數量前四多的主詞(%d,%d,%d,%d)", priority[0], priority[1], priority[2], priority[3]);*/
  181.  
  182. 							if (prefered != 4 && enumerate == '2') {
  183. 								for (a = 1; a <= 6; a++) {
  184. 									if (core[l].skills[a])
  185. 										selected[a] -= 1;
  186. 								}
  187. 								continue;
  188. 							}
  189. 							printf("\n");
  190.  
  191. 							for (a = 0; a < 4; a++) {
  192. 								printf("核心編號:%d, [ ", hold[a] + 1);
  193.  
  194. 								for (b = 0; b < 3; b++) {
  195.  
  196. 									printf("%d ", core[hold[a]].skillOrder[b]);
  197.  
  198.  
  199. 								}
  200. 								printf("]\n");
  201.  
  202. 							}	
  203.  
  204. 							if (enumerate == '2'&&prefered == 4) { return; }
  205. 							else if(enumerate == '3') { return; }
  206. 						}
  207.  
  208. 					}
  209. 					for (a = 1; a <= 6; a++) {
  210. 						if (core[l].skills[a])
  211. 							selected[a] -= 1;
  212. 					}
  213. 				}
  214. 				for (a = 1; a <= 6; a++) {
  215. 					if (core[k].skills[a])
  216. 						selected[a] -= 1;
  217. 				}
  218. 			}
  219. 			for (a = 1; a <= 6; a++) {
  220. 				if (core[j].skills[a])
  221. 					selected[a] -= 1;
  222. 			}
  223. 		}
  224. 		for (a = 1; a <= 6; a++) {
  225. 			if (core[i].skills[a])
  226. 				selected[a] -= 1;
  227. 		}
  228. 	}
  229. 	printf("\ndone(若都沒有出現\"找到一組4核六技!\"字樣，表示沒有發現可能的組合，請再多蒐集一些核心!)\n");
  230.  
  231.  
  232.  }
  233. void modeTwo(struct Core core[], int coreCount, int enumerate, int priority[6]) {
  234. 	/*
  235. 	finding core combinations using brute force loops
  236. 	*/
  237. 	int selected[10] = { 0 };
  238. 	int hold[6];
  239. 	int i, j, k, l,m,n, a, b, sentinel;
  240. 	printf("start\n");
  241. 	for (i = 0; i < coreCount; i++) {
  242. 		for (a = 1; a <= 9; a++) {
  243. 			if (core[i].skills[a])
  244. 				selected[a] += 1;
  245. 		}
  246. 		hold[0] = i;//selected 1 core
  247. 		for (j = 0; j < coreCount; j++) {
  248. 			if (j == i)continue;
  249. 			for (a = 1; a <= 9; a++) {
  250. 				if (core[j].skills[a])
  251. 					selected[a] += 1;
  252. 			}
  253. 			hold[1] = j;//selected 2 cores
  254. 			for (k = 0; k < coreCount; k++) {
  255. 				if (k == i || k == j)continue;
  256. 				sentinel = 0;
  257. 				for (a = 1; a <= 9; a++) {
  258. 					if (core[k].skills[a] && 3 == selected[a] + 1)/*if 3== selected[a] + 1, then reject subtree*/
  259. 					{
  260. 						sentinel = 1; break;
  261. 					}
  262.  
  263. 				}
  264. 				if (sentinel)continue;
  265. 				for (a = 1; a <= 9; a++) {
  266. 					if (core[k].skills[a])
  267. 						selected[a] += 1;
  268. 				}
  269.  
  270. 				hold[2] = k;//selected 3 cores
  271. 				for (l = 0; l < coreCount; l++) {
  272. 					if (l == i || l == j||l==k)continue;
  273. 					sentinel = 0;
  274. 					for (a = 1; a <= 9; a++) {
  275. 						if (core[l].skills[a] && 3 == selected[a] + 1)/*if 3== selected[a] + 1, then reject subtree*/
  276. 						{
  277. 							sentinel = 1; break;
  278. 						}
  279.  
  280. 					}
  281. 					if (sentinel)continue;
  282. 					for (a = 1; a <= 9; a++) {
  283. 						if (core[l].skills[a])
  284. 							selected[a] += 1;
  285. 					}
  286.  
  287. 					hold[3] = l;//selected 4 cores
  288. 					for (m = 0; m < coreCount; m++) {
  289. 						if (m == i || m == j || m == k||m==l)continue;
  290. 						sentinel = 0;
  291. 						for (a = 1; a <= 9; a++) {
  292. 							if (core[m].skills[a] && 3 == selected[a] + 1)/*if 3== selected[a] + 1, then reject subtree*/
  293. 							{
  294. 								sentinel = 1; break;
  295. 							}
  296.  
  297. 						}
  298. 						if (sentinel)continue;
  299. 						for (a = 1; a <= 9; a++) {
  300. 							if (core[m].skills[a])
  301. 								selected[a] += 1;
  302. 						}
  303.  
  304. 						hold[4] = m;//selected 5 cores
  305. 						for (n = 0; n < coreCount; n++) {
  306. 							if (n == i || n == j || n == k||n==l||n==m)continue;
  307. 							for (a = 1; a <= 9; a++) {
  308. 								if (core[n].skills[a])
  309. 									selected[a] += 1;
  310. 							}
  311. 							hold[5] = n;//selected 6 cores
  312. 							if (selected[1] == 2/*final check*/
  313. 								&& selected[2] == 2
  314. 								&& selected[3] == 2
  315. 								&& selected[4] == 2
  316. 								&& selected[5] == 2
  317. 								&& selected[6] == 2
  318. 								&& selected[7] == 2
  319. 								&& selected[8] == 2
  320. 								&& selected[9] == 2
  321. 								) {
  322. 								//printf("\n���@��6��9��!\n");
  323. 								int checkDup[6] = { 0 }, dupCount = 0;
  324. 								for (a = 0; a < 6; a++) {
  325. 									//	printf("�֤߽s��:%d, [ ", hold[a] + 1);
  326. 									checkDup[dupCount] = core[hold[a]].skillOrder[2];
  327. 									dupCount++;
  328.  
  329. 									//	printf("]\n\n");
  330.  
  331. 								}
  332. 								if (checkDup[0] != checkDup[1] &&
  333. 									checkDup[0] != checkDup[2] &&
  334. 									checkDup[0] != checkDup[3] &&
  335. 									checkDup[0] != checkDup[4] &&
  336. 									checkDup[0] != checkDup[5] &&
  337. 									checkDup[1] != checkDup[2] &&
  338. 									checkDup[1] != checkDup[3] &&
  339. 									checkDup[1] != checkDup[4] &&
  340. 									checkDup[1] != checkDup[5] &&
  341. 									checkDup[2] != checkDup[3] &&
  342. 									checkDup[2] != checkDup[4] &&
  343. 									checkDup[2] != checkDup[5] &&
  344. 									checkDup[3] != checkDup[4] &&
  345. 									checkDup[3] != checkDup[5] &&
  346. 									checkDup[4] != checkDup[5]
  347. 									) {
  348. 									int prefered = 0;
  349. 									/*compare priority and checkdup*/
  350.  
  351. 									for (a = 0; a < 6; a++)
  352. 										for (b = 0; b < 6; b++)
  353. 											if (priority[a] == checkDup[b])
  354. 												prefered++;
  355. 									if (prefered == 6)
  356. 										printf("\n���@�դ��֤E��!�A�B�����D���ŦX�z�����ƶq�e���h���D��!\n");
  357. 									/*else
  358. 										printf("\n���@�դ��֤E��!�A�������D�����ŦX�z�����ƶq�e�|�h���D��");
  359. 									printf("[%d,%d,%d,", priority[0], priority[1], priority[2]);
  360. 									printf("%d,%d,%d]", priority[3], priority[4], priority[5]);*/
  361.  
  362. 									if (prefered != 6 && enumerate == '2') {/*unselect 6th core*/ 
  363. 										for (a = 1; a <= 9; a++) {
  364. 											if (core[n].skills[a])
  365. 												selected[a] -= 1;
  366. 										}
  367. 										continue;
  368. 									}
  369. 									for (a = 0; a < 6; a++) {
  370. 										printf("�֤߽s��:%d, [ ", hold[a] + 1);
  371.  
  372. 										for (b = 0; b < 3; b++) {
  373.  
  374. 											printf("%d ", core[hold[a]].skillOrder[b]);
  375.  
  376.  
  377. 										}
  378. 										printf("]\n");
  379.  
  380. 									}
  381.  
  382. 									if (enumerate == '2'&&prefered == 6) { return; }
  383. 									else if (enumerate == '3') { return; }
  384. 								}
  385. 							}
  386. 							for (a = 1; a <= 9; a++) {/*unselect 6th core*/
  387. 								if (core[n].skills[a])
  388. 									selected[a] -= 1;
  389. 							}
  390. 						}
  391. 						for (a = 1; a <= 9; a++) {//unselect 5th core
  392. 							if (core[m].skills[a])
  393. 								selected[a] -= 1;
  394. 						}
  395. 					}
  396. 					for (a = 1; a <= 9; a++) {/*unselect 4th core*/
  397. 						if (core[l].skills[a])
  398. 							selected[a] -= 1;
  399. 					}
  400. 				}
  401. 				for (a = 1; a <= 9; a++) {/*unselect 3rd core*/
  402. 					if (core[k].skills[a])
  403. 						selected[a] -= 1;
  404. 				}
  405.  
  406. 			}
  407. 			for (a = 1; a <= 9; a++) {/*unselect 2nd core*/
  408. 				if (core[j].skills[a])
  409. 					selected[a] -= 1;
  410. 			}
  411. 		}
  412. 		for (a = 1; a <= 9; a++) {/*unselect 1st core*/
  413. 			if (core[i].skills[a])
  414. 				selected[a] -= 1;
  415. 		}
  416. 	}
  417. 	printf("\ndone(若都沒有出現\"找到一組6核9技!\"字樣，表示沒有發現可能的組合，請再多蒐集一些核心!)\n");
  418.  
  419. }
Success #stdin #stdout 0s 5320KB
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stdin
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Standard input is empty
stdout
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�֤߽s��:2, [ 1 4 3 ]
�֤߽s��:3, [ 1 2 5 ]
�֤߽s��:4, [ 1 6 4 ]
�֤߽s��:5, [ 1 3 6 ]
�֤߽s��:6, [ 4 2 1 ]
�֤߽s��:7, [ 4 3 5 ]
�֤߽s��:8, [ 4 1 5 ]
�֤߽s��:9, [ 4 5 1 ]
�֤߽s��:10, [ 4 3 5 ]
�֤߽s��:11, [ 4 3 1 ]
�֤߽s��:12, [ 4 1 2 ]
�֤߽s��:13, [ 4 1 2 ]
�֤߽s��:14, [ 4 3 2 ]
�֤߽s��:15, [ 4 1 3 ]
�֤߽s��:16, [ 4 5 1 ]
�֤߽s��:17, [ 6 5 2 ]
�֤߽s��:18, [ 6 5 1 ]
�֤߽s��:19, [ 6 2 1 ]
�֤߽s��:20, [ 6 1 5 ]
�֤߽s��:21, [ 6 3 2 ]
�֤߽s��:22, [ 6 2 4 ]
�֤߽s��:23, [ 6 2 3 ]
�֤߽s��:24, [ 6 1 5 ]
�֤߽s��:25, [ 6 5 3 ]
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�֤߽s��:27, [ 5 4 6 ]
�֤߽s��:28, [ 5 6 3 ]
�֤߽s��:29, [ 5 6 2 ]
�֤߽s��:30, [ 5 4 2 ]
�֤߽s��:31, [ 3 6 1 ]
�֤߽s��:32, [ 3 5 2 ]
�֤߽s��:33, [ 3 6 2 ]
�֤߽s��:34, [ 3 2 6 ]
�֤߽s��:35, [ 3 5 6 ]
�֤߽s��:36, [ 3 4 6 ]
�֤߽s��:37, [ 3 6 4 ]
�֤߽s��:38, [ 3 2 4 ]
�֤߽s��:39, [ 3 5 4 ]
�֤߽s��:40, [ 3 6 5 ]
�֤߽s��:41, [ 2 6 4 ]
�֤߽s��:42, [ 2 1 3 ]
�֤߽s��:43, [ 2 4 5 ]
�֤߽s��:44, [ 2 3 6 ]
�֤߽s��:45, [ 2 1 4 ]
�C�X�D���X�{����
�D��7�X�{0
�D��8�X�{0
�D��9�X�{0
�D��6�X�{6
�D��1�X�{7
�D��4�X�{7
�D��3�X�{8
�D��5�X�{8
�D��2�X�{9
start

找到一組四核六技!，且它的主詞符合您持有數量前四多的主詞!(2,5,3,4)

核心編號:1, [ 1 2 3 ]
核心編號:4, [ 1 6 4 ]
核心編號:7, [ 4 3 5 ]
核心編號:17, [ 6 5 2 ]
https://ideone.com/gUNSE6
language:
C++ (gcc 8.3)
created:
10 hours ago
visibility:
public

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