程序运行失败的回文子串数

Question

I have a homework assignment that asks us to use dynamic programming to count the number of possible palindrome substrings in a given string. My program, theoretically, should work, but when I run it, the run fails every time. I do not get any compiler errors. Here's my code:

#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

/*
 * 
 */
/*bool isPalindrome(string input){
    bool isPal = true;
    int j = (input.length() - 1);
    for(int i = 0; i < (input.length()/2); i++){
        if(input[i] != input[j]){
            isPal = false;
        }
        j--;
    }
    return isPal;
}*/

bool isPalindrome(string input, bool array[5000][5000], int i, int j) {
    int a = i;
    int b = j;

    if(array[i][j])
        return true;

    while(input[a] == input[b]){
        array[a][b] = true;
        a++;
        b--;
        if((b - a) <= 2){
            break;
        }
    }
    return array[i][j];
}

int numPalindrome(string input) {
    bool array[5000][5000];
    for(int k = 0; k < 5000; k++)
        for(int n = 0; n < 5000; n++)
            array[k][n] = false;

    int count = 0;
    for(int i = 0; i < input.length(); i++){
        for(int j = i; j <= input.length(); j++){
            if(isPalindrome(input, array, i, j)){
                count++;
            }
        }
    }
    return count;
}

int main(int argc, char** argv) {
    int count = numPalindrome("xabcba");
    cout << "Count: " << count << endl;
    return 0;
}

Can anyone help find out why my code will not run? Thanks.

Answer 1

All palindromes in a string can be represented as a graph like this:

len
 4                    abba
                     /  | \
 3        aba       /   |  \       aca
         / | \     /    |   \     / | \
 2      /  |  \   /    bb    \   /  |  \   aa
       /   |   \ /    /  \    \ /   |   \ /  \
 1    a    b    a    b    b    a    c    a     a

This relation-ship can be used pretty easy to find all palindromes and reducing space-complexity to O(n) . And reducing the runtime-complexity by a great ammount, since we only need to search for all palindromes with length 2 and 3. So unless the input is worstcase ("aaaaaa...aaaaaa" for eg.), we can reduce the number of searched words by a great amount. The basic idea is to simply store the position of palindromes at an even or uneven position and search for palindromes that are two characters longer. Given a list of the positions of palindromes with length n , we can search for palindromes with length n + 2 by simply checking if the character next to the previous palindrome on the left and right side are equal.