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// C++ program palindromic tree by khalid anwar
#include "bits/stdc++.h"
using namespace std;
#define MAXN 1000
struct Node
{
// store start and end indexes of current
// Node inclusively
int start, end;
// stores length of substring
int length;
// stores insertion Node for all characters a-z
int insertEdg[26];
// stores the Maximum Palindromic Suffix Node for
// the current Node
int suffixEdg;
};
// two special dummy Nodes as explained above
Node root1, root2;
// stores Node information for constant time access
Node tree[MAXN];
// Keeps track the current Node while insertion
int currNode;
string s;
int ptr;
void
insert (int idx)
{
//STEP 1//
/* search for Node X such that s[idx] X S[idx]
is maximum palindrome ending at position idx
iterate down the suffix link of currNode to
find X */
int tmp = currNode;
while (true)
{
int curLength = tree[tmp].length;
if (idx - curLength >= 1 and s[idx] == s[idx - curLength - 1])
break;
tmp = tree[tmp].suffixEdg;
}
/* Now we have found X ....
* X = string at Node tmp
* Check : if s[idx] X s[idx] already exists or not*/
if (tree[tmp].insertEdg[s[idx] - 'a'] != 0)
{
// s[idx] X s[idx] already exists in the tree
currNode = tree[tmp].insertEdg[s[idx] - 'a'];
return;
}
// creating new Node
ptr++;
// making new Node as child of X with
// weight as s[idx]
tree[tmp].insertEdg[s[idx] - 'a'] = ptr;
// calculating length of new Node
tree[ptr].length = tree[tmp].length + 2;
// updating end point for new Node
tree[ptr].end = idx;
// updating the start for new Node
tree[ptr].start = idx - tree[ptr].length + 1;
//STEP 2//
/* Setting the suffix edge for the newly created
Node tree[ptr]. Finding some String Y such that
s[idx] + Y + s[idx] is longest possible
palindromic suffix for newly created Node. */
tmp = tree[tmp].suffixEdg;
// making new Node as current Node
currNode = ptr;
if (tree[currNode].length == 1)
{
// if new palindrome's length is 1
// making its suffix link to be null string
tree[currNode].suffixEdg = 2;
return;
}
while (true)
{
int curLength = tree[tmp].length;
if (idx - curLength >= 1 and s[idx] == s[idx - curLength - 1])
break;
tmp = tree[tmp].suffixEdg;
}
// Now we have found string Y
// linking current Nodes suffix link with s[idx]+Y+s[idx]
tree[currNode].suffixEdg = tree[tmp].insertEdg[s[idx] - 'a'];
}
// driver program
int
main ()
{
// initializing the tree
root1.length = -1;
root1.suffixEdg = 1;
root2.length = 0;
root2.suffixEdg = 1;
tree[1] = root1;
tree[2] = root2;
ptr = 2;
currNode = 1;
// given string
s = "challenge";
int l = s.length ();
for (int i = 0; i < l; i++)
insert (i);
// printing all of its distinct palindromic
// substring
cout << "All distinct palindromic substring for "
<<s << " : \n";
for (int i = 3; i <= ptr; i++)
{
cout << i - 2 << ") ";
for (int j = tree[i].start; j <= tree[i].end; j++)
cout << s[j];
cout << endl;
}
return 0;
}
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