A. Road to Cinema

很明显满足二分性质的题目。

题意：某人在起点处，到终点的距离为s。 汽车租赁公司提供n中车型，每种车型有属性ci（租车费用），vi（油箱容量）。 车子有两种前进方式 ：①. 慢速：1km消耗1L汽油，花费2分钟。 ②.快速：1km消耗2L汽油，花费1分钟。 路上有k个加油站，加油不需要花费时间，且直接给油箱加满。 问在t分钟内到达终点的最小花费是多少？（租车子的费用）  若无法到达终点，输出-1

不谈二分的写法.. 我们考虑离散化可修改的点  和限制的点位置   如果发现有限制就结束。那么思路很简单了。一个数组存分段的最大耗油量。ans = 跑最快情况下的最小时间的最大值。依次牺牲多余时间减小耗油即可。取个max就可以了。

算上排序的时间依旧是klogk+n。不好写坑点多。。

开了流量写CF.... 500MB没了......  我R   算了算了不写....

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