POJ 1743 Musical Theme (the suffix array is used to find the longest repeated substring that cannot be overlapped)

Question: give some notes and find the longest recurring rotation length.

 

From the meaning of the question, we can know that as long as the adjacent difference is equal, we can create an array that is not adjacent. The question requires the longest length of the compound substring, and cannot overlap.

Because the adjacent difference may be negative, add 100 in a uniform way and convert it to the number between 0 and.

Then obtain the height array, which indicates the longest common prefix of the adjacent suffix.

Second answer, and then determine

You can group the Suffixes by height. If the height is greater than the length, you need to determine whether the suffixes overlap. Use the sa difference value. For details, see the code.

[Cpp]
# Include <iostream>
# Include <cstdio>
# Include <cstring>
# Include <queue>
# Include <cmath>
# Include <string>
# Include <vector>
# Include <algorithm>
# Include <map>
# Include <set>
# Define maxn50005
# Define eps 1e-8
# Define zero (a) fabs (a) <eps
Using namespace std;
// Obtain the suffix Array Using the multiplication algorithm.
Int wa [maxn], wb [maxn], wv [maxn], Ws [maxn];
Int cmp (int * r, int a, int B, int l)
{Return r [a] = r [B] & r [a + l] = r [B + l];}
Void da (const int * r, int * sa, int n, int m ){
Int I, j, p, * x = wa, * y = wb, * t;
For (I = 0; I <m; I ++) Ws [I] = 0;
For (I = 0; I <n; I ++) Ws [x [I] = r [I] ++;
For (I = 1; I <m; I ++) Ws [I] + = Ws [I-1];
For (I = n-1; I> = 0; I --) sa [-- Ws [x [I] = I;
For (j = 1, p = 1; p <n; j * = 2, m = p ){
For (p = 0, I = n-j; 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 ++) Ws [I] = 0;
For (I = 0; I <n; I ++) Ws [wv [I] ++;
For (I = 1; I <m; I ++) Ws [I] + = Ws [I-1];
For (I = n-1; I> = 0; I --) sa [-- Ws [wv [I] = y [I];
For (t = x, x = y, y = t, p = 1, x [sa [0] = 0, I = 1; I <n; I ++)
X [sa [I] = cmp (y, sa [I-1], sa [I], j )? P-1: p ++;
}
Return;
}
Int sa [maxn], Rank [maxn], height [maxn];
// Calculate the height Array
Void calheight (const int * r, int * sa, int n ){
Int I, j, k = 0;
For (I = 1; I <= n; I ++) Rank [sa [I] = I;
For (I = 0; I <n; height [Rank [I ++] = k)
For (k? K --: 0, j = sa [Rank [I]-1]; r [I + k] = r [j + k]; k ++ );
Return;
}
Int n;
Int a [maxn], B [maxn];
Bool check (int m ){
Int mmax = 0, mmin = n;
For (int I = 1; I <= n; I ++ ){
If (height [I] <m ){
Mmax = sa [I];
Mmin = sa [I];
}
Else {
Mmax = max (mmax, max (sa [I], sa [I-1]);
Mmin = min (mmin, min (sa [I], sa [I-1]);
If (mmax-mmin> m) return true;
}
}
Return false;
}
Int main (){
While (scanf ("% d", & n )! = EOF & n ){
For (int I = 0; I <n; I ++) scanf ("% d", & a [I]);
For (int I = 0; I <n-1; I ++) B [I] = a [I + 1]-a [I] + 100;
N --;
B [n] = 0;
Da (B, sa, n + 1,200 );
Calheight (B, sa, n );
If (! Check (4 )){
Printf ("0 \ n ");
Continue;
}
Int ans, low = 0, high = n, mid;
While (low <= high ){
Mid = (low + high)/2;
If (check (mid )){
Ans = mid;
Low = mid + 1;
}
Else
High = mid-1;
}
Printf ("% d \ n", ans + 1 );
}
Return 0;