相关链接
题目传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4059
神犇题解Ⅰ：http://blog.csdn.net/popoqqq/article/details/46380617
神犇题解Ⅱ：http://krydom.com/bzoj4059/
解题报告
考虑直接找出所有不无聊的区间
直接 $ O(n^2)$ 枚举的话，会T掉，于是考虑分治
假如当前考虑到的区间为(l,r)，且这个区间中一个只出现过一次的数为x，下标为p
那么显然，横跨x的区间合法，两边的区间就递归下去做就好啦！
但如果每一次我们从左到右扫描x的话，时间复杂度长这样：$ T(n) = T(n – 1) + n$
这样的话，上限是 $ O(n^2)$ 显然不清真
考虑同时从两边开始扫描的话，复杂度就变为了： $ T(n) = T(p) + T(n – p) + \min (n – p,p)$
于是这个是：$ O(n\log n)$ 的，于是就可以愉快的暴力啦！
至于复杂度的证明的话，我们发现这货的形式和线段树的启发式合并的式子长得一毛一样
因为线段树的启发式合并是 $ O(n\log n)$ 的，那么这货的复杂度也就是对的啦！
当然你也可以使用归纳法什么的来证明复杂度
甚至你可以根本不用这个暴力的算法，而使用优雅的矩形面积并来做：
http://www.cnblogs.com/DaD3zZ-Beyonder/p/5803964.html
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