I and OI
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摘要: const maxn=100; maxm=trunc(ln(maxn)/ln(2))+1;var h,sa,rank,tmp,tax:array[0..maxn*2]of longint; f:array[0..maxn,0..maxm] of longint; ... 阅读全文
posted @ 2011-09-25 17:23 exponent 阅读(616) 评论(0) 推荐(0) 编辑
摘要: 母串:S 子串:T extend[i]=LCP(S[i..n],T) next[i]=LCP(T[i..n],T) 假设extend[1..k]已计算出,现计算extend[k+1] 令P=max{I+extend[I]-1} (I=1..k) 并令使P取最大值的I为a 则有 S[a..p]=T[1... 阅读全文
posted @ 2011-09-25 17:22 exponent 阅读(253) 评论(0) 推荐(0) 编辑