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#include <iostream>
#include <fstream>
#include <cstring>
#include <vector>
using namespace std;
ifstream fin("strmatch.in");
ofstream fout("strmatch.out");
const int MAXN = 2000005;
string Pattern, Text, S;
int z[MAXN];
int Left, Right;
vector<int> Solution;
int main() {
fin >> Pattern >> Text;
S = Pattern + Text;
int SLength = S.size();
int PatternLength = Pattern.size();
for (int k = 1; k < SLength; ++k) {
if (Right < k) {//we're outside of the Z-box
if (S[0] == S[k]) {//first letters equal
Left = Right = k;
for (int i = 0; S[i] == S[Right]; ++i)
++Right;
--Right;
z[k] = Right - k + 1;
}
}else {//inside the Z-box
/*if the length of the Z-box starting at
(k - Left + 1) (aka the position of letter k in the Pattern)
is STRICTLY smaller than the current Z-box (between Left and Right)
then we can use the information as we know we cannot go any further
with the comparisons
*/
if (z[k - Left] < Right - k + 1) {
z[k] = z[k - Left];
}else
//here we know we might find merged strings, so we
//go on with the comparisons beyond Right, actually extending out Z-box
if (z[k - Left] >= Right - k + 1) {
//0 1 2 3 4 5 6 7 8 9 10 11 12 13
//A B A C A B B C A B A B A B
int q = Right + 1;
for (int i = z[Right - Left]; S[i] == S[q]; ++i) {
++q;
}
z[k] = q - k;
}
}
}
/*for (int i = 0; i < SLength; ++i) {
cerr << z[i] << ' ';
}*/
for (int i = PatternLength; i < SLength; ++i)
if (z[i] == PatternLength)
Solution.push_back(i - PatternLength);
fout << Solution.size() << '\n';
for (size_t i = 0; i < Solution.size(); ++i)
fout << Solution[i] << ' ';
fin.close();
fout.close();
return 0;
}