题目大意是给出n个数,求能否用这n个数组成两个和相等的队列,如果能则输出最大可能的单个队列数字和。
一开始是想着用0/1背包来做的,不过后来发现有bug,虽然x和x/2都有可能取到,但x/2却未必有两份(x表示队列和)。看了相关讨论才想到可以用两个队列的差值进行DP,用f[i,j]表示取到第i个数,队列差为j时的单个队列和最大值,最后输出f[n,0]即可。因为f[i]只与f[i-1]有关,故可以用滚动数组优化空间复杂度。另外有一个效果不错的优化是可以在DP前先将n个数升序排序,这样可以减少DP过程中的枚举量。
3024745 2008-08-06 16:19:51 Accepted 2059 FPC 00:00.21 424K IwfWcf@LZOI
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