Light OJ 1258 – Making Huge Palindromes


/*
Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1258
*/

#include <bits/stdc++.h>

#define pii pair <int,int>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define oo 1<<29
#define dd double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define MAX 1000007
#define CIN ios_base::sync_with_stdio(0)
#define SZ(a) (int)a.size()
#define getint(a) scanf("%d",&a)
#define getint2(a,b) scanf("%d%d",&a,&b)
#define getint3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define loop(i,n) for(int i=0;i<n;i++)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d: ",z)
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define inf 1000000
#define rtintlim 46340
#define llim 9223372036854775808
#define rtllim 3037000499
#define ull unsigned long long
#define I int

using namespace std;

/* Bits operation */
int Set(int n,int pos)  { return n = n | 1<<pos;}
bool check(int n,int pos) { return n & 1<<pos;}
int Reset(int n, int pos) { return n=n & ~(1<<pos);}
/*----------------*/

int lps[MAX];

void computePrefixFunction(string& P)
{
    lps[0]=-1;
    int k=-1;
    int n=SZ(P);
    for(int i=1;i<n;i++)
    {
        while(k>-1 && P[i]!=P[k+1])
            k=lps[k];
        if(P[i]==P[k+1])
            k++;
        lps[i]=k;
    }
}

 int KMP(string& T, string& P)
 {
     int n=SZ(T);
     int m=SZ(P);
     int k=-1;
     computePrefixFunction(P);
     int cnt=0;
     for(int i=0;i<n;i++)
     {
         while(k>-1 && T[i]!=P[k+1])
            k=lps[k];
         if(T[i]==P[k+1])
            k++;
         if(i==n-1)
            cnt=k;
     }
     return cnt+1;
 }
int main()
{
    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int t;
    getint(t);
    TEST_CASE(t)
    {
        string txt,pattarn;
        cin>>txt;
        pattarn=txt;
        reverse(all(pattarn));
        int n=KMP(txt,pattarn);
        int m=SZ(txt);
        PRINT_CASE;
        cout<<m+m-n<<endl;
    }
    return 0;
}

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Muhammad Mahir Hasan Chowdhury

what does kmp do in this problem and why you make your ans by doing (m+m-n)?