May not overlap the longest repeated substring

Topic links: https://cn.vjudge.net/contest/318888#overview

 

The meaning of problems: Given a piano tone sequence P [Values ​​range (1 to 88)], is now required to find a sequence satisfying

1, a length of at least 5

2, the sequence can transpose, i.e., the presence of two sub-sequences, one sub-sequence satisfies the plus / minus a number of other sequence can be

3, the sequence can not have two intersecting portion.

Meaning of the questions is simply looking for duplicate substring longest non-overlapping

 

Ideas:

In fact, this question may overlap and maximum repeat request longest substring like. We then find the longest base may overlap as long as the longest substring repeated again to determine what the sa [i] coincides with it so as not

 

  1 #include <stdio.h>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <string.h>
  5 #include <stdlib.h>
  6 #include <math.h>
  7 #include <queue>
  8 #include <set>
  9 
 10 #define INF 0x3f3f3f3f
 11 #define pii pair<int,int>
 12 #define LL long long
 13 using namespace std;
 14 typedef unsigned long long ull;
 15 const int maxn = 2e6+6;
 16 
 17 int s[maxn];
 18 int sa[maxn],t[maxn],t2[maxn],c[maxn];
 19 int Rank[maxn],height[maxn];
 20 
 21 void build_sa(int n,int m)
 22 {
 23     int i,j,*x=t,*y=t2;
 24     for (i=0;i<m;i++)
 25         c[i] = 0;
 26     for (i=0;i<n;i++)
 27         c[x[i] = s[i]]++;
 28     for (i=1;i<m;i++)
 29         c[i] += c[i-1];
 30     for (i=n-1;i>=0;i--)
 31         sa[--c[x[i]]] = i;
 32     for (int k=1;k<=n;k<<=1)
 33     {
 34         int p = 0;
 35         for (i=n-k;i<n;i++)
 36             y[p++] = i;
 37         for (i=0;i<n;i++)
 38         {
 39             if (sa[i]>=k)
 40                 y[p++] = sa[i]-k;
 41         }
 42         for (i=0;i<m;i++)
 43             c[i] = 0;
 44         for (i=0;i<n;i++)
 45             c[x[y[i]]]++;
 46         for (i=1;i<m;i++)
 47             c[i] += c[i-1];
 48         for (i=n-1;i>=0;i--)
 49             sa[--c[x[y[i]]]] = y[i];
 50         swap(x,y);
 51         p = 1;
 52         x[sa[0]] = 0;
 53         for (i=1;i<n;i++)
 54             x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
 55         if (p>=n)
 56             break;
 57         m = p;
 58     }
 59 }
 60 
 61 
 62 void getHeight(int n){
 63     int i,j,k=0;
 64     for (i=1;i<=n;i++){
 65         Rank[sa[i]] = i;
 66     }
 67     for (i=0;i<n;i++){
 68         if (k)
 69             k--;
 70         j = sa[Rank[i]-1];
 71         while (s[i+k] == s[j+k])
 72             k++;
 73         height[Rank[i]] = k;
 74     }
 75 }
 76 
 77 int n;
 78 bool check(int k){
 79     int maxsa=sa[0],minsa=sa[0];
 80     for (int i=1;i<n;i++){
 81         if (height[i]>=k){
 82             minsa = min(sa[i],minsa);
 83             maxsa =max (in [i], maxsa);
84          }
 85          else {
 86              if (maxsa to time> = k)
 87                  return  true ;
88              maxsa = a [i];
89              minsa = in [i];
90          }
 91      }
 92      if (maxsa to time> = k) {
 93          return  true ;
94      }
 95      return  false ;
96  }
 97  
98  int main () {
 99      seconds (~scanf("%d",&n) && n!=0){
100         for (int i=0;i<n;i++){
101             scanf("%d",&s[i]);
102         }
103         for (int i=0;i<n;i++){
104             if (i == n-1)
105                 s[i] = 0;
106             else
107                 s[i] = s[i+1]-s[i] + 100;
108         }
109         build_sa(n,200);
110         getHeight(n-1);
111         int l=0,r=n,mid;
112         while (l<=r){
113             mid = (l+r)>>1;
114             if (check(mid)){
115                 l = mid+1;
116             }
117             else{
118                 r = mid-1;
119             }
120         }
121         if (r<4){
122             printf("0\n");
123         }
124         else{
125             printf("%d\n",r+1);
126         }
127     }
128     return 0;
129 }