Pagini recente » Arhiva de probleme, pregatire pentru concursuri de informatica | Cod sursa (job #1266797) | Cod sursa (job #1090096) | Cod sursa (job #2490119) | Cod sursa (job #1851047)
#include <cstdio>
#include <cstring>
int n;
int pos[1005], lps[2000005];
char pat[2000005], txt[2000005];
// Fills lps[] for given patttern pat[0..M-1]
void computeLPSArray(const char *pat, int M, int *lps)
{
// length of the previous longest prefix suffix
int len = 0;
lps[0] = 0; // lps[0] is always 0
// the loop calculates lps[i] for i = 1 to M-1
int i = 1;
while (i < M)
{
if (pat[i] == pat[len])
{
len++;
lps[i] = len;
i++;
}
else // (pat[i] != pat[len])
{
// This is tricky. Consider the example.
// AAACAAAA and i = 7. The idea is similar
// to search step.
if (len != 0)
{
len = lps[len-1];
// Also, note that we do not increment
// i here
}
else // if (len == 0)
{
lps[i] = 0;
i++;
}
}
}
}
// Prints occurrences of txt[] in pat[]
void KMPSearch(const char *pat, const char *txt)
{
int M = strlen(pat);
int N = strlen(txt);
// create lps[] that will hold the longest prefix suffix
// values for pattern
// int lps[M];
// Preprocess the pattern (calculate lps[] array)
computeLPSArray(pat, M, lps);
int i = 0; // index for txt[]
int j = 0; // index for pat[]
while (i < N)
{
if (pat[j] == txt[i])
{
j++;
i++;
}
if (j == M)
{
if (n<1000)
pos[n++] = i-j;
j = lps[j-1];
}
// mismatch after j matches
else if (i < N && pat[j] != txt[i])
{
// Do not match lps[0..lps[j-1]] characters,
// they will match anyway
if (j != 0)
j = lps[j-1];
else
i = i+1;
}
}
}
// Driver program to test above function
int main()
{
freopen("strmatch.in", "r", stdin);
freopen("strmatch.out", "w", stdout);
scanf("%s", pat);
scanf("%s", txt);
KMPSearch(pat, txt);
printf("%d\n", n);
for (int i = 0; i < n; ++i)
printf("%d ", pos[i]);
return 0;
}