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平衡树,297ms水过……

Posted by Ever_ljq at 2011-06-13 22:19:23 on Problem 3277
快排 + 扫描。

扫描时用平衡树维护当前尚存的线段的高度,遇右端点删除对应左端点,
每次ans += tree[root].max * (point[i].position - ps[i-1].postion)。

输出我用的cout。

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