## Algoritmo de SuffixArray.

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``````/**
* Author: ???
* Source: PDF UPC
* Description: string matching
* Time: \$O(P + T)\$ where \$P\$ is the length of the pattern, \$T\$ is length of the text
* Status: Tested
*/
typedef pair<int, int> ii;

#define MAX_N 100010 // O(n log n)
char T[MAX_N]; // up to 100K characters
int n; //length of string
int RA[MAX_N], tempRA[MAX_N];
int SA[MAX_N], tempSA[MAX_N];
int c[MAX_N];
char P[MAX_N]; // the pattern string (for string matching)
int m; // the length of pattern string
int Phi[MAX_N];// for computing LCP
int PLCP[MAX_N];
int LCP[MAX_N];
bool cmp(int a, int b) { return strcmp(T + a, T + b) < 0; }

void countingSort(int k) {  // O(n)
int i, sum, maxi = max(300, n); // up to 255 chars
memset(c, 0, sizeof c);
for (i = 0; i < n; i++)
c[i + k < n ? RA[i + k] : 0]++;
for (i = sum = 0; i < maxi; i++) {
int t = c[i]; c[i] = sum; sum += t;
}
for (i = 0; i < n; i++) tempSA[c[SA[i]+k < n ? RA[SA[i]+k] : 0]++] = SA[i];
for (i = 0; i < n; i++) SA[i] = tempSA[i];
}
void constructSA() { // Up to 100000 characters
int i, k, r;
for (i = 0; i < n; i++) RA[i] = T[i];
for (i = 0; i < n; i++) SA[i] = i;
for (k = 1; k < n; k <<= 1) {
countingSort(k);
countingSort(0);
tempRA[SA[0]] = r = 0;
for (i = 1; i < n; i++)
tempRA[SA[i]] =
(RA[SA[i]] == RA[SA[i-1]] && RA[SA[i]+k] == RA[SA[i-1]+k]) ? r : ++r;
for (i = 0; i < n; i++)
RA[i] = tempRA[i];
if (RA[SA[n-1]] == n-1) break;
} }

void computeLCP() {
int i, L;
Phi[SA[0]] = -1;
for (i = 1; i < n; i++)
Phi[SA[i]] = SA[i-1];
for (i = L = 0; i < n; i++) {// LCP in O(n)
if (Phi[i] == -1) { PLCP[i] = 0; continue; }
while (T[i + L] == T[Phi[i] + L]) L++;
PLCP[i] = L;
L = max(L-1, 0);
}
for (i = 0; i < n; i++)
LCP[i] = PLCP[SA[i]];
}

ii stringMatching() { // O(m log n)
int lo = 0, hi = n-1, mid = lo;// valid matching = [0..n-1]
while (lo < hi) {
mid = (lo + hi) / 2;
int res = strncmp(T + SA[mid], P, m);
if (res >= 0) hi = mid;
else          lo = mid + 1;
}
if (strncmp(T + SA[lo], P, m) != 0) return ii(-1, -1); // if not found
ii ans; ans.first = lo;
lo = 0; hi = n - 1; mid = lo;
while (lo < hi) {
mid = (lo + hi) / 2;
int res = strncmp(T + SA[mid], P, m);
if (res > 0) hi = mid;
else         lo = mid + 1;
}
if (strncmp(T + SA[hi], P, m) != 0) hi--;
ans.second = hi;
return ans;
}
ii LRS() {
int i, idx = 0, maxLCP = -1;
for (i = 1; i < n; i++) // O(n)
if (LCP[i] > maxLCP)
maxLCP = LCP[i], idx = i;
return ii(maxLCP, idx);
}
int owner(int idx) { return (idx < n-m-1) ? 1 : 2; }

ii LCS() {
int i, idx = 0, maxLCP = -1;
for (i = 1; i < n; i++) // O(n)
if (owner(SA[i]) != owner(SA[i-1]) && LCP[i] > maxLCP)
maxLCP = LCP[i], idx = i;
return ii(maxLCP, idx);
}
int main() {
printf("Suffix Array:\n");
strcpy(T, "GATAGACA");
n = (int)strlen(T);
T[n++] = '\$';
//if '\n' uncomment T[n-1] = '\$'; T[n] = 0;
constructSA();// O(n log n)
printf("\nSR of string T = '%s':\n", T);
printf(" i\tSA[i]\tSuffix\n");
for (int i = 0; i < n; i++) printf("%2d\t%2d\t%s\n", i, SA[i], T + SA[i]);
computeLCP();// O(n)
ii ans = LRS();
char lrsans[MAX_N];
strncpy(lrsans, T + SA[ans.second], ans.first);
printf("\nLongest Repeated Substring O(n)\n");
printf("\nLRS is '%s' = %d\n\n", lrsans, ans.first);
printf("\nString matching O(m log n)\n");
//printf("\nenter string P below,find P in T:\n");
strcpy(P, "A");
m = (int)strlen(P);
//if '\n' uncomment P[m-1] = 0; m--;
ii pos = stringMatching();
if (pos.first != -1 && pos.second != -1) {
printf("%s is found SA[%d..%d] of %s\n", P, pos.first, pos.second, T);
printf("They are:\n");
for (int i = pos.first; i <= pos.second; i++)
printf("  %s\n", T + SA[i]);
printf("\nLongest Common Substring O(n)\n");
printf("\nRemember, T = '%s'\nNow, enter another string P:\n", T);
// T already has '\$' at the back
strcpy(P, "CATA");
m = (int)strlen(P);
// if '\n' is read, uncomment the next line
//P[m-1] = 0; m--;
strcat(T, P);
strcat(T, "#");
n = (int)strlen(T);
//Un prefijo de un sufijo es un substring
constructSA();
computeLCP();
printf("\nLongest Common Prefix O(n)\n");
printf("\nThe LCP information of 'T+P' = '%s':\n", T);
printf(" i\tSA[i]\tLCP[i]\tOwner\tSuffix\n");
for (int i = 0; i < n; i++)
printf("%2d\t%2d\t%2d\t%2d\t%s\n", i, SA[i], LCP[i], owner(SA[i]), T + SA[i]);
ans = LCS();
char lcsans[MAX_N];
strncpy(lcsans, T + SA[ans.second], ans.first);
printf("\nThe LCS is '%s' = %d\n", lcsans, ans.first);
return 0;
}``````