Aizu 1370 Hidden Anagrams 字符串哈希

Hidden Anagrams

Time Limit : 10 sec, Memory Limit : 262144 KB

Problem D Hidden Anagrams

An anagram is a word or a phrase that is formed by rearranging the letters of another. For instance, by rearranging the letters of "William Shakespeare," we can have its anagrams "I am a weakish speller," "I'll make a wise phrase," and so on. Note that when AA is an anagram of BB, BB is an anagram of AA.

In the above examples, differences in letter cases are ignored, and word spaces and punctuation symbols are freely inserted and/or removed. These rules are common but not applied here; only exact matching of the letters is considered.

For two strings s1s1 and s2s2 of letters, if a substring s′1s1′ of s1s1 is an anagram of a substring s′2s2′ of s2s2, we call s′1s1′a hidden anagram of the two strings, s1s1 and s2s2. Of course, s′2s2′ is also a hidden anagram of them.

Your task is to write a program that, for given two strings, computes the length of the longest hidden anagrams of them.

Suppose, for instance, that "anagram" and "grandmother" are given. Their substrings "nagr" and "gran" are hidden anagrams since by moving letters you can have one from the other. They are the longest since any substrings of "grandmother" of lengths five or more must contain "d" or "o" that "anagram" does not. In this case, therefore, the length of the longest hidden anagrams is four. Note that a substring must be a sequence of letters occurring consecutively in the original string and so "nagrm" and "granm" are not hidden anagrams.

Input

The input consists of a single test case in two lines.

s1s1
s2s2

s1s1 and s2s2 are strings consisting of lowercase letters (a through z) and their lengths are between 1 and 4000, inclusive.

Output

Output the length of the longest hidden anagrams of s1s1 and s2s2. If there are no hidden anagrams, print a zero.

Sample Input 1

anagramgrandmother

Sample Output 1

4

Sample Input 2

williamshakespeareiamaweakishspeller

Sample Output 2

18

Sample Input 3

aaaaaaaabbbbbbbbxxxxxabababxxxxxabab

Sample Output 3

6

Sample Input 4

abababacdcdcdefefefghghghghgh

Sample Output 4

0

给两个长度最多4000的字符串，判断它们的所有子串中，每个字母出现次数相同的子串的最大长度。

从刘大爷那里学到了一个奇技淫巧，同时也十分玄学的算法：字符串哈希。

对于这题，统计子串中每个字母出现的次数，算出哈希函数之后用map判重即可。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <bitset>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef unsigned long long ull;typedef long double ld;const int maxn=5005,inf=0x3f3f3f3f;char s[maxn],t[maxn];int a[26];ull seed1=26,seed2=1e9+7;ull hash() {ull sum=seed1;for (int i=0;i<26;i++) {sum*=seed2;sum+=a[i];}return sum;}int main() {map<ull,int> HashTable;scanf("%s",s);scanf("%s",t);int lena,lenb,i,j,k,ans;lena=strlen(s),lenb=strlen(t);int len=min(lena,lenb);ans=0;for (i=1;i<=len;i++) {HashTable.clear();mem0(a);for (j=0;j<i;j++) a[s[j]-'a']++;for (j=0;j<lena-i;j++) {HashTable[hash()]++;a[s[j]-'a']--;a[s[j+i]-'a']++;}//cout << hash() << endl;HashTable[hash()]++;mem0(a);for (j=0;j<i;j++) a[t[j]-'a']++;for (j=0;j<lenb-i;j++) {//cout << hash() << endl;if (HashTable[hash()]) ans=max(ans,i);a[t[j]-'a']--;a[t[j+i]-'a']++;}if (HashTable[hash()]) ans=max(ans,i);}printf("%d\n",ans);return 0;}