{"action":"closed","issue":{"url":"https://api.github.com/repos/RWebRTC/RWebRTC/issues/71","repository_url":"https://api.github.com/repos/RWebRTC/RWebRTC","labels_url":"https://api.github.com/repos/RWebRTC/RWebRTC/issues/71/labels{/name}","comments_url":"https://api.github.com/repos/RWebRTC/RWebRTC/issues/71/comments","events_url":"https://api.github.com/repos/RWebRTC/RWebRTC/issues/71/events","html_url":"https://github.com/RWebRTC/RWebRTC/issues/71","id":181805162,"number":71,"title":"棋盘覆盖","user":{"login":"RWebRTC","id":21374839,"avatar_url":"https://avatars.githubusercontent.com/u/21374839?v=3","gravatar_id":"","url":"https://api.github.com/users/RWebRTC","html_url":"https://github.com/RWebRTC","followers_url":"https://api.github.com/users/RWebRTC/followers","following_url":"https://api.github.com/users/RWebRTC/following{/other_user}","gists_url":"https://api.github.com/users/RWebRTC/gists{/gist_id}","starred_url":"https://api.github.com/users/RWebRTC/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/RWebRTC/subscriptions","organizations_url":"https://api.github.com/users/RWebRTC/orgs","repos_url":"https://api.github.com/users/RWebRTC/repos","events_url":"https://api.github.com/users/RWebRTC/events{/privacy}","received_events_url":"https://api.github.com/users/RWebRTC/received_events","type":"User","site_admin":false},"labels":[{"url":"https://api.github.com/repos/RWebRTC/RWebRTC/labels/%E7%AE%97%E6%B3%95%E5%AF%BC%E8%AE%BA","name":"算法导论","color":"b60205"}],"state":"closed","locked":false,"assignee":null,"assignees":[],"milestone":null,"comments":0,"created_at":"2016-10-08T05:07:01Z","updated_at":"2016-10-19T13:23:01Z","closed_at":"2016-10-19T13:23:01Z","body":"### 描述\r\n\r\n在一个2^k * 2^k个方格组成的棋盘中，有一个方格与其它的不同，若使用以下四种L型骨牌覆盖除这个特殊方格的其它方格，如何覆盖。\r\n\r\n![b](https://cloud.githubusercontent.com/assets/21374839/19519607/4419a8dc-963f-11e6-9ecf-c80bd32e79c1.gif)\r\n\r\n![a](https://cloud.githubusercontent.com/assets/21374839/19519575/20c1976e-963f-11e6-97eb-e6b2838005f2.gif)\r\n\r\n### 思路\r\n\r\n实现的基本原理是将2^k * 2^k的棋盘分成四块2^(k - 1) * 2^(k - 1)的子棋盘，特殊方格一定在其中的一个子棋盘中，如果特殊方格在某一个子棋盘中，继续递归处理这个子棋盘，直到这个子棋盘中只有一个方格为止如果特殊方 格不在某一个子棋盘中，将这个子棋盘中的相应的位置设为骨牌号，将这个无特殊方格的了棋盘转换为有特殊方格的子棋盘，然后再递归处理这个子棋盘。以上原理 如图3所示。\r\n\r\n![c](https://cloud.githubusercontent.com/assets/21374839/19519593/3499074a-963f-11e6-872c-8a9decf24fb7.gif)\r\n\r\n```c++\r\nvoid deal(){\r\n    // 四个 if else\r\n    if(在左上角){\r\n        // 递归处理\r\n    }else{\r\n        // 左上角的右下角设置成特殊块，递归处理\r\n    }\r\n    // 同理\r\n    return;\r\n}\r\n```"}}
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