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在兩個串中唯一出現的最小公共子串

D. Match & Catch time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output

Police headquarter is monitoring signal on different frequency levels. They have got two suspiciously encoded strings?s1?and?s2?from two different frequencies as signals. They are suspecting that these two strings are from two different criminals and they are planning to do some evil task.

Now they are trying to find a common substring of minimum length between these two strings. The substring must occur only once in the first string, and also it must occur only once in the second string.

Given two strings?s1?and?s2?consist of lowercase Latin letters, find the smallest (by length) common substring?p?of both?s1?and?s2, wherep?is a unique substring in?s1?and also in?s2. See notes for formal definition of substring and uniqueness.

Input

The first line of input contains?s1?and the second line contains?s2?(1?≤?|s1|,?|s2|?≤?5000). Both strings consist of lowercase Latin letters.

Output

Print the length of the smallest common unique substring of?s1?and?s2. If there are no common unique substrings of?s1?and?s2?print -1.

Sample test(s) input apple pepperoni output 2 input lover driver output 1 input bidhan roy output -1 input testsetses teeptes output 3 Note

Imagine we have string?a?=?a1a2a3...a|a|, where?|a|?is the length of string?a, and?ai?is the?ith?letter of the string.

We will call string?alal?+?1al?+?2...ar?(1?≤?l?≤?r?≤?|a|)?the substring?[l,?r]?of the string?a.

The substring?[l,?r]?is unique in?a?if and only if there is no pair?l1,?r1?such that?l1?≠?l?and the substring?[l1,?r1]?is equal to the substring[l,?r]?in?a.

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int maxn=10100,INF=0x3f3f3f3f;int sa[maxn],rank[maxn],rank2[maxn],h[maxn],c[maxn],*x,*y,ans[maxn]; char str[maxn];bool cmp(int*r,int a,int b,int l,int n) {if(r[a]==r[b]&&a+l<n&&b+l<n&&r[a+l]==r[b+l]) return true;return false; }bool radix_sort(int n,int sz) {for(int i=0;i<sz;i++) c[i]=0;for(int i=0;i<n;i++) c[x[y[i]]]++;for(int i=1;i<sz;i++) c[i]+=c[i-1];for(int i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i]; }void get_sa(char c[],int n,int sz=128) {x=rank,y=rank2;for(int i=0;i<n;i++) x[i]=c[i],y[i]=i;radix_sort(n,sz);for(int len=1;len<n;len*=2){int yid=0;for(int i=n-len;i<n;i++) y[yid++]=i;for(int i=0;i<n;i++) if(sa[i]>=len) y[yid++]=sa[i]-len;radix_sort(n,sz);swap(x,y);x[sa[0]]=yid=0;for(int i=1;i<n;i++){x[sa[i]]=cmp(y,sa[i],sa[i-1],len,n)?yid:++yid;}sz=yid+1;if(sz>=n) break;}for(int i=0;i<n;i++) rank[i]=x[i]; }void get_h(char str[],int n) {int k=0; h[0]=0;for(int i=0;i<n;i++){if(rank[i]==0) continue;k=max(k-1,0);int j=sa[rank[i]-1];while(i+k<n&&j+k<n&&str[i+k]==str[j+k]) k++;h[rank[i]]=k;} }int main() {cin>>str;int sg=strlen(str);str[sg]=127;cin>>str+sg+1;int n=strlen(str);get_sa(str,n);get_h(str,n);int ans=INF;int s1=0,s2=0,last=-1;for(int i=1;i<n;i++){if(sa[i-1]<sg&&sa[i]<sg) continue;if(sa[i-1]>sg&&sa[i]>sg) continue;int pre=h[i-1];int next=h[i+1];if(h[i]>max(pre,next)){ans=min(ans,max(pre,next)+1);}}if(ans==INF) ans=-1;printf("%d\n",ans);return 0; }

轉載于:https://www.cnblogs.com/mengfanrong/p/3763720.html

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