c_cpp 使用后缀数组计算给定文本中子字符串的出现次数

Posted 2021-05-14

/*====================================================================
 Title: Computes Longest Repeated Substring from a given string
 Complexity: O(n.log(n))
 Author : Sudipto Chandra (Dipu)
 =====================================================================*/
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a, b, sizeof(a))
#define loop(i, x) for(__typeof((x).begin()) i=(x).begin(); i!=(x).end(); ++i)
#define rloop(i, x) for(__typeof((x).rbegin()) i=(x).rbegin(); i!=(x).rend(); ++i)
/*------------------------------------------------------------------------------------*/

int test, cas = 1;

const int SIZ = 10000050; // maximum possible size

int n; // text length
int m; // pattern length
char T[SIZ]; // text
char P[SIZ]; // pattern
int SA[SIZ], tempSA[SIZ]; // the sorted suffixes
int RA[SIZ], tempRA[SIZ]; // ranks of suffix array
int L[SIZ]; // used in counting sort

inline int getRA(int i)
{
    return (i < n) ? RA[i] : 0;
}

void radix_sort(int k)
{
    mem(L, 0);
    // count frequencies
    for(int i = 0; i < n; ++i)
    {
        L[getRA(i + k)]++;
    }
    // save first index of every characters
    int mx = max(n, 130);
    for(int i = 0, s = 0; i < mx; ++i)
    {
        int x = L[i];
        L[i] = s;
        s += x;
    }
    // build sorted tempSA
    for(int i = 0; i < n; ++i)
    {
        int& x = L[getRA(SA[i] + k)];
        tempSA[x++] = SA[i];
    }
    // copy tempSA to SA
    for(int i = 0; i < n; ++i)
    {
        SA[i] = tempSA[i];
    }
}
// text must ends with a $
void buildSA()
{
    // initialize
    n = strlen(T);
    T[n++] = '$', T[n] = 0; // append $
    for(int i = 0; i < n; ++i)
    {
        SA[i] = i;
        RA[i] = T[i];
    }

    // algorithm loop
    for(int k = 1; k < n; k <<= 1)
    {
        // sort by k-th ranks
        radix_sort(k);
        radix_sort(0);
        // compute new ranks
        tempRA[SA[0]] = 0;
        for(int i = 1, r = 0; i < n; ++i)
        {
            if(getRA(SA[i-1]) != getRA(SA[i])) {
                r++;
            }
            else if(getRA(SA[i-1]+k) != getRA(SA[i]+k)) {
                r++;
            }
            tempRA[SA[i]] = r;
        }
        for(int i = 0; i < n; ++i)
        {
            RA[i] = tempRA[i];
        }
        if(RA[SA[n - 1]] == n - 1) break;
    }
}

// Count the number of times a given pattern P appears
// in the string T. Complexity: O(n.log(n))
int countOccurences()
{
    m = strlen(P);
    SA[n] = n;

    int first, last;
    int low, high, mid;

    // find lower bound
    low = 0, high = n;
    while(low < high)
    {
        mid = (low + high) >> 1;
        printf("lower: %d %d %d\n", low, mid, high);
        if(strncmp(P, T + SA[mid], m) <= 0)
        {
            high = mid;
        }
        else
        {
            low = mid + 1;
        }
    }
    printf("lower: %d %d %d\n", low, mid, high);
    first = low;

    // find higher bound
    low = 0, high = n;
    while(low < high)
    {
        mid = (low + high) >> 1;
        printf("upper: %d %d %d\n", low, mid, high);
        if(strncmp(P, T + SA[mid], m) < 0)
        {
            high = mid;
        }
        else
        {
            low = mid + 1;
        }
    }
    printf("upper: %d %d %d\n", low, mid, high);
    last = low;

    printf("%d %d\n", first, last);
    return last - first;
}

int main()
{
    /*
        GATAGACA
        GA

        abcabxabcd
        abc

        mississippi
        i
    */

    printf("Text: ");
    scanf("%s", T);
    printf("Pattern: ");
    scanf("%s", P);

    buildSA();

    for(int i = 0; i < n; ++i)
    {
        printf("%2d | %2d | %s\n", i, SA[i], T + SA[i]);
    }

    int ans = countOccurences();

    printf("Pattern found %d times\n", ans);

    return 0;
}