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我的简洁算法

Posted by stream_speed at 2006-08-04 16:38:48 on Problem 2923
从tot/(C1+C2)到n枚举,最少tot/(C1+C2)趟,最多n趟,从小到达搜索,找到答案就退出,比如搜索在i趟内能不能完工,可以转换为在一趟内用两种型号的车各i辆能不能完工,然后用深搜就可以实现了,很简洁

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