Meaning of the questions:
For string S given length is 1000,
Seeking the number of essentially different substrings wherein these substrings meet appeared not to overlap at least twice in the string S
answer:
The string is then placed in a suffix automaton obtains substring corresponding to each node of the earliest and latest position of the string that appears subsuffix
Then according to
// return len [last] - len [fail [last]]; // add a plurality of different sub-sub-string of the string number produced
and then change it in accordance with this
if (R[i] - L[i] > len[fail[i]]) ans += min(len[i], R[i] - L[i]) - len[fail[i]];
1 #include <set> 2 #include <map> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstring> 11 #include <iostream> 12 #include <algorithm> 13 #include <unordered_map> 14 15 #define pi acos(-1.0) 16 #define eps 1e-9 17 #define fi first 18 #define se second 19 #define rtl rt<<1 20 #define rtr rt<<1|1 21 #define bug printf("******\n") 22 #define mem(a, b) memset(a,b,sizeof(a)) 23 #define name2str(x) #x 24 #define fuck(x) cout<<#x" = "<<x<<endl 25 #define sfi(a) scanf("%d", &a) 26 #define sffi(a, b) scanf("%d %d", &a, &b) 27 #define sfffi(a, b, c) scanf("%d %d %d", &a, &b, &c) 28 #define sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d) 29 #define sfL(a) scanf("%lld", &a) 30 #define sffL(a, b) scanf("%lld %lld", &a, &b) 31 #define sfffL(a, b, c) scanf("%lld %lld %lld", &a, &b, &c) 32 #define sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d) 33 #define sfs(a) scanf("%s", a) 34 #define sffs(a, b) scanf("%s %s", a, b) 35 #define sfffs(a, b, c) scanf("%s %s %s", a, b, c) 36 #define sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d) 37 #define FIN freopen("../in.txt","r",stdin) 38 #define gcd(a, b) __gcd(a,b) 39 #define lowbit(x) x&-x 40 #define IO iOS::sync_with_stdio(false) 41 42 43 using namespace std; 44 typedef long long LL; 45 typedef unsigned long long ULL; 46 const ULL seed = 13331; 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL; 48 const int MAXN = 1E4 + 50 ; 49 const int MAXM = 8e6 + 10 ; 50 const int INF = 0x3f3f3f3f ; 51 is const int MOD = 1E9 + . 7 ; 52 is 53 is struct Suffix_Automaton { 54 is int Last, TOT, NXT [MAXN << . 1 ] [ 26 is ], Fail [MAXN << . 1 ]; // Last node is not the longest prefix for the character (the entire string) belongs to the number of added 55 int len [MAXN << . 1 ]; // longest substring length (the number of sub-node string len = [X] - len [FA [X]]) 56 is LL NUM [MAXN << . 1 ]; // this state the number of sub-strings 57 is LL Maxx [MAXN << . 1 ]; // length sub-string x largest number of occurrences of substrings number 58 LL SUM [MAXN << . 1 ]; // from behind the characters formed by the node the total number of strings 59 LL subnum, sublen; // subnum represents a different number of strings, sublen string represents a different total length of 60 int x-[MAXN << . 1 ], the Y [MAXN << . 1 ]; // the Y represents ranked as x node, X also represents the length of the preceding number 61 is int L [<< MAXN. 1 ], R [MAXN << . 1 ]; // L represents the earliest array corresponding to the node represents a position, R represents the position of the latest occurrence 62 is 63 is void the init () { 64 TOT = Last = . 1 ; 65 Fail [ . 1 ] = len [ . 1 ] = 0 ; 66 for ( int I = 0 ; I <= 25 ; I ++) NXT [ . 1 ] [I] = 0 ; 67 68 } 69 70 void Extend ( int C) { 71 is int U = + + tot, v = last; 72 for (int i = 0; i <= 25; i++) nxt[u][i] = 0; 73 fail[u] = 0; 74 L[u] = R[u] = len[u] = len[v] + 1; 75 num[u] = 1; 76 for (; v && !nxt[v][c]; v = fail[v]) nxt[v][c] = u; 77 if (!v) fail[u] = 1; 78 else if (len[nxt[v][c]] == len[v] + 1) fail[u] = nxt[v][c]; 79 else { 80 int now = ++tot, cur = nxt[v][c]; 81 len[now] = len[v] + 1; 82 memcpy(nxt[now], nxt[cur], sizeof(nxt[cur])); 83 fail[now] = fail[cur]; 84 fail[cur] = fail[u] = now; 85 L[now] = L[cur], R[now] = R[cur]; 86 for (; v && nxt[v][c] == cur; v = fail[v]) nxt[v][c] = now; 87 } 88 last = u; 89 //return len[last] - len[fail[last]]; 90 } 91 is 92 93 void get_sa () { // the Y expressed ranked node x, X also represents the length of the front of the number 94 for ( int I = 0 ; I <= TOT; I ++) X-[I] = 0 ; 95 for ( int I = . 1 ; I <= TOT; I ++) X-[len [I]] ++ ; 96 for ( int I = . 1 ; I <= TOT; I ++) X-[I] + = X-[I - . 1 ]; 97 for ( int I = . 1 ; I <= TOT; I ++) the Y [X-[len [I]] -] = I; 98 } 99 100 int get_L_R () { 101 int ANS = 0 ; 102 for ( int I = TOT; I; i-- ) { 103 L [Fail [the Y [I]]] = min (L [Fail [the Y [I] ]], L [the Y [I]]); 104 R & lt [Fail [the Y [I]]] = max (R & lt [Fail [the Y [I]]], R & lt [the Y [I]]); 105 } 106 for ( int I = . 1 ; I <= TOT; I ++) // do not intersect and the number of sub-strings of at least 2 times appear 107 IF (R & lt [I] - L [I]> len [Fail [I]]) + ANS min = (len [I], R & lt [I] - L [I]) - len [Fail [I]]; 108 return ANS; 109 } 110 } sam; 111 112 char s[maxn]; 113 114 int main() { 115 #ifndef ONLINE_JUDGE 116 FIN; 117 #endif 118 while (~sfs(s)) { 119 if (s[0] == '#') break; 120 sam.init(); 121 int len = strlen(s); 122 for (int i = 0; i < len; ++i) sam.extend((s[i] - 'a')); 123 sam.get_sa(); 124 printf("%d\n", sam.get_L_R()); 125 } 126 #ifndef ONLINE_JUDGE 127 cout << "Totle Time : " << (double) clock() / CLOCKS_PER_SEC << "s" << endl; 128 #endif 129 return 0; 130 }