UVa 11475 Extend to Palindrome

方法：kmp

其实这道题算是一个字符串匹配，原string 为 str， 把str 反转后得到 rev，看rev在str什么位置可以匹配到，多出的部分不算。匹配的方法有很多种，这道题用kmp比较容易。也可以用suffix array 或者 hash。

kmp 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 <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);


string str, rev;
void getfail(const string &str, int *f)
{
    int m = str.length();
    f[0] = f[1] = 0;
    for (int i = 1; i < m; ++i)
    {
        int j = f[i];
        while (j && str[i] != str[j]) j = f[j];
        f[i+1] = str[i]==str[j]?j+1:0;
    }
}
int f[100001];
int main()
{
    while (cin >> str)
    {
        rev = str;
        reverse(rev.begin(), rev.end());
        getfail(rev, f);
        int ans;
        int n = str.length();
        int j = 0;
        for (int i = 0; i < n; ++i)
        {
            while (j && rev[j] != str[i]) j = f[j];
            if (rev[j] == str[i]) ++j;
            //cerr << str.substr(0,i+1) << " " << j << newline;
        }
        cout << str << rev.substr(j) << newline;
    }
}