UVa 11888 Abnormal 89's

方法：Manacher

Manacher算法在O(length) 时间内求出各个回文子串的长度。O(length) 时间检查时那一种情况。

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);


const int maxn = 2e5+1;
char ma[maxn*2];
int p[maxn*2];
int l;
void Manacher(const string &str)
{
    int len = str.length();
    l = 0;
    ma[l++] = '$';
    ma[l++] = '#';
    rep(i, len)
    {
        ma[l++] = str[i];
        ma[l++] = '#';
    }
    ma[l] = 0;
    int pnow = 0, pid = 0;
    for (int i = 1; i < l; ++i)
    {
        if (pnow > i) p[i] = min(p[2*pid-i], pnow-i);
        else p[i] = 1;
        for (; ma[i-p[i]] == ma[i+p[i]]; ++p[i]);
        if (i+p[i] > pnow)
        {
            pnow = i+p[i];  pid = i;
        }
    }
}
/*
 abaaba
 .  ,  a  ,  b  ,  a  ,  a  ,  b  ,  a  ,
 0  1  2  1  4  1  2  7  2  1  4  1  2  1
 0  1  2  3  4  5  6  7  8  9 10 11 12 13
 */
int main()
{
    int T;  cin >> T;
    repn(kase, T)
    {
        string str;
        cin >> str;
        Manacher(str);

        bool done = false;
        for (int i = 2; i < (1+l)/2; ++i)
            if (p[i] == i && p[(i+p[i]-1+l)/2] == l-(i+p[i]-1+l)/2)
            {
                cout << "alindrome
";
                done = true;
                break;
            }
        if (!done)
            if (p[(l+1)/2]-1 == str.length())
                cout << "palindrome
";
            else
            cout << "simple
";

    }
}