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高级搜索专题

·1 min
ACM 算法竞赛
Note: This article is available in Chinese only. 本文暂无英文版本。 View original

基础的搜索BFS和DFS,自己找题切吧…

高级搜索的题集就在下面,自己看着办吧…

努力爆搜,努力剪枝吧~~~

【Level 1】

HDOJ-1429 胜利大逃亡(续)　 HDOJ-1885 Key Task HDOJ-1226 超级密码 HDOJ-1664 Different Digits HDOJ-2821 Pusher HDOJ-2128 Tempter of the Bone II HDOJ-3533 Escape HDOJ-4101 Ali and Baba HDOJ-3839 Ancient Messages HDOJ-1685 Booksort HDOJ-2614 Beat HDOJ-3309 Roll The Cube HDOJ-1067 Gap HDOJ-2181 哈密顿绕行世界问题 HDOJ-2437 Jerboas HDOJ-2102 A计划 HDOJ-1195 Open the Lock HDOJ-3295 An interesting mobile game HDOJ-2259 Continuous Same Game(2) HDOJ-3681 Prison Break HDOJ-3085 Nightmare Ⅱ

【Level 2】 POJ-1475 Pushing Boxes POJ-3635 Full Tank? POJ-2044 Weather Forecast POJ-2449 Remmarguts’ Date POJ-1324 Holedox Moving POJ-3322 Bloxorz I POJ-2308 Dearboy’s Puzzle POJ-2688 Cleaning Robot POJ-1376 Robot POJ-1190 生日蛋糕 POJ-1184 聪明的打字员 HDOJ-4012 Paint on a Wall HDOJ-3766 Knight’s Trip HDOJ-2605 Snake HDOJ-3121 FreeOpen HDOJ-3900 Unblock Me HDOJ-1732 Push Box HDOJ-2913 Traveling Cube HDOJ-3001 Travelling HDOJ-4090 GemAnd Prince

【Level 3】 HDOJ-1401 Solitaire HDOJ-4127 Flood-it! HDOJ-1560 DNA sequence HDOJ-2808 Islands HDOJ-1430 魔板 HDOJ-1043 Eight HDOJ-3567 Eight II HDOJ-1667 The Rotation Game HDOJ-2234 无题I HDOJ-1813 Escape from Tetris HDOJ-2918 Tobo or not Tobo HDOJ-3459 Rubik 2×2×2 HDOJ-2953 Rubiks Cube ZOJ-2477 Magic Cube HDOJ-2691 2-Dimensional Rubik’s Cube HDOJ-2467 Deja vu HDOJ-2485 Destroying the bus stations

Related

【2-SAT问题】（转自kuangbin的博客）

·5 mins
ACM 算法竞赛
【2-SAT问题】 现有一个由N个布尔值组成的序列A，给出一些限制关系，比如A[x] AND A[y]=0、A[x] OR A[y] OR A[z]=1等，要确定A[0..N-1]的值，使得其满足所有限制关系。这个称为SAT问题，特别的，若每种限制关系中最多只对两个元素进行限制，则称为2-SAT问题。

斯特林数

·1 min
ACM 算法竞赛
http://baike.baidu.com/link?url=nsN1-rcs3Gs0jNurWLSDk6AJ9jmhl_3pfkQmYK7vZoe7BsoTij48Si3It9XeNM4uA7gST-1ITQsAx0bv5si9_q

cf 570 C. Replacement (暴力)

·2 mins
ACM 算法竞赛
C. Replacement time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Daniel has a string s, consisting of lowercase English letters and period signs (characters ‘.’). Let’s define the operation of replacement as the following sequence of steps: find a substring “..” (two consecutive periods) in string s, of all occurrences of the substring let’s choose the first one, and replace this substring with string “.”. In other words, during the replacement operation, the first two consecutive periods are replaced by one. If string s contains no two consecutive periods, then nothing happens.