Hihocoder #1415: suffix array three • Repeat Melody 3

#1415: suffix array three • Repeat Melody 3Time limit:5000mscase time limit:1000msmemory limit:256mb description

Small hi usually a big hobby is playing the piano. We know that a musical melody is represented as a series of numbers consisting of N in length. Little hi has practiced a lot of music and found that many of the melodies in the works have a common part.

Melody is a continuous series, if the same melody in works A and b in the same time, this melody is a and b common parts, such as in Abab in Bababab and Cabacababc have appeared. Little hi want to know what is the longest common melody of the two works?

Tips on how to solve problems

Input

A total of two lines. A single line of string that contains only lowercase letters. The string length does not exceed 100000.

Output

An integer line that represents the answer.

Sample Input

Abcdefgabacabca

Sample Output

3

Analysis:

Connecting two strings together becomes the longest common prefix for suffixes, but since the prefix cannot span two strings, we add a character after the first string that does not appear, then the second string, and then the maximum height of all the non-identical string suffixes ...

Code:

#include <algorithm> #include <iostream> #include <cstring> #include <cstdio>//by neighthornusing namespace Std;//zhao kan fei niao mu fei hui,ying chuang hua luo lian chui diconst int maxn=200000+5;int n , T,s[maxn],gs[maxn],sa[maxn],wb[maxn],wv[maxn],ran[maxn],height[maxn];char s1[maxn],s2[maxn];inline BOOL cmp (INT * X,int a,int b,int L) {return x[a]==x[b]&&x[a+l]==x[b+l];} inline void da (int *sa,int *x,int n,int m) {int i,j,p,*y=wb;for (i=0;i<m;i++) gs[i]=0;for (i=0;i<n;i++) gs[x[i]]++; for (i=1;i<m;i++) gs[i]+=gs[i-1];for (i=n-1;~i;i--) sa[--gs[x[i]]]=i;for (j=1,p=1;p<n;j<<=1,m=p) {for (i= n-j,p=0;i<n;i++) y[p++]=i;for (i=0;i<n;i++) if (sa[i]>=j) y[p++]=sa[i]-j;for (i=0;i<n;i++) wv[i]=x[y[i]]; for (i=0;i<m;i++) gs[i]=0;for (i=0;i<n;i++) gs[wv[i]]++;for (i=1;i<m;i++) gs[i]+=gs[i-1];for (i=n-1;~i;i--) Sa[--gs[wv[i]]]=y[i];p =1;swap (x, y), x[sa[0]]=0;for (i=1;i<n;i++) x[sa[i]]=cmp (y,sa[i],sa[i-1],j) p-1:p++;}} inline void calheight (int n) {int i,j, K=0;for (i=0;i<=n;i++) ran[sa[i]]=i;for (i=0;i<n;height[ran[i++]]=k) for (k?k--:233,j=sa[ran[i]-1];s[i+k]==s [j+k];k++);} inline int solve (int n) {int ans=0;for (int i=1;i<=n;i++) if (sa[i]<t&&sa[i-1]>t) | | (sa[i]>t&&sa[i-1]<t)) Ans=max (Ans,height[i]); return ans;} Signed main (void) {scanf ("%s%s", S1,s2), T=n=strlen (S1); for (int i=0;i<n;i++) s[i]=s1[i]-' a ' +1;s[n]=100;int lala= strlen (S2); for (int i=1;i<=lala;i++) s[i+n]=s2[i-1]-' a ' +1;n+=lala+1;for (int i=0;i<n;i++) ran[i]=s[i];d a (SA, RAN,N+1,101); Calheight (n);p rintf ("%d\n", Solve (n)); return 0;} The Cap ou pas cap. Pas Cap.

　　

by Neighthorn

Hihocoder #1415: suffix array three • Repeat Melody 3