bzoj3942

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【BZOJ3940】【BZOJ3942】[Usaco2015 Feb]Censoring AC自动机/KMP/hash+栈

[BZOJ3942][Usaco2015 Feb]Censoring Description Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of material to read while waiting around in the barn during milking sessions. Unfortunately, the…

bzoj3940&&bzoj3942 Ac自动机||kpm算法

方法就是维护一个动态栈记录栈的每一位匹配到串的哪一位的编号第一道kmp第二道ac自动机自己理会 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ; char stack[M],s[M],t[M],f[M],next[M]; int len,top; void getfail(){ ;i<len;i++){ int j=f[i]; while(j&&…

BZOJ3942 [Usaco2015 Feb]Censoring

维护一个栈...如果栈顶出现了要被删除的字符串就全删掉就好了,判断的话...kmp就行了 /************************************************************** Problem: 3942 User: rausen Language: C++ Result: Accepted Time:224 ms Memory:11548 kb *******************************************************…

bzoj3942——2016——3——15

题目大意: 3942: [Usaco2015 Feb]Censoring Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 220 Solved: 115[Submit][Status][Discuss] Description Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of ma…

BZOJ3942: [Usaco2015 Feb]Censoring 栈+KMP

Description Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of material to read while waiting around in the barn during milking sessions. Unfortunately, the latest issue contains a rather inap…

Bzoj3942 Censoring(KMP)

\(KMP\)问题的核心在于数组\(next\)(或者\(pre\)/\(fail\),各种叫法),几乎所有的此类型题都是需要计算\(next\)的. 这里解释一波\(next\):即满足字符子串\(s[1...k]=s[j-k+1...j]\)的最大\(k\).比如说: 字符串\(abaca\)的\(next[5]\)就是\(1\),因为只有\(s[1]=s[5]\). 而字符串\(abcab\)的\(next[5]\)就是2,同理,\(s[1...2]=s[4...5]\). 弄清楚了这一点…