[hdu3068 最长回文]Manacher算法，O(N)求最长回文子串

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#pragma comment(linker, "/STACK:10240000")
#include <bits/stdc++.h>
using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;

#ifndef ONLINE_JUDGE
namespace Debug {
void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
}
#endif // ONLINE_JUDGE
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
/* -------------------------------------------------------------------------------- */

const int maxn = 3e5 + 7;

/** 求字符串每个位置的最大回文半径，在字符串中找最长回文子串 **/
struct Manacher {
    int p[maxn];/** 回文半径 **/
    char s[maxn];
    void init(char str[]) {
        strcpy(s, str);
        int n = strlen(s);
        s[n * 2 + 1] = 0;
        for (int i = n * 2; i; i -= 2) {
            s[i] = ‘#‘;
            s[i - 1] = s[i / 2 - 1];
        }
        s[0] = ‘#‘;
    }
    /** 求每个点的最大回文半径 **/
    void work() {
        int r = 0, id = 0;
        p[0] = 1;
        for (int i = 1; s[i]; i ++) {
            p[i] = i <= r? min(r - i + 1, p[2 * id - i]) : 1;
            if (p[i] >= r - i + 1) {
                r = (id = i) + p[i] - 1;
                while (2 * i - r - 1 >= 0 && s[r + 1] == s[2 * i - r - 1]) {
                    r ++;
                    p[i] ++;
                }
            }
        }
    }
    /** 求最长回文串的长度 **/
    int solve() {
        work();
        int ans = 1;
        for (int i = 0; s[i]; i ++) {
            ans = max(ans, p[i] - 1);
        }
        return ans;
    }
};
Manacher solver;

char s[maxn];

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    while (~scanf("%s", s)) {
        solver.init(s);
        printf("%d\n", solver.solve());
    }
    return 0;
}