HDU3294 manacher

来源：互联网发布：python 异或运算符编辑：程序博客网时间：2024/04/29 09:56

题意：将字符串按一定规则变换，输出变换后的字符串中的最长回文串的起始和终止位置，并输出回文串。
题意：manacher模板题

#include <iostream>#include <cstring>using namespace std;const int MAXN = 200000 + 10;char c, s[MAXN * 2];int p[MAXN * 2];int st, t, maxlen;void Turn(char* s){    int len = strlen(s);    int dis = c - 'a';    for(int i = 0; i < len; i++)    {        if(s[i] - dis >= 'a') s[i] -= dis;        else        {            s[i] = 'z' - (dis - (s[i] - 'a') - 1);        }    }}void Manacher(char* s){    memset(p, 0, sizeof(p));    maxlen = 0;    int len = strlen(s);    int id = 0;    //²åÈë#    for(int i = len; i >= 0; i--)    {        s[2 * i + 2] = s[i];        s[2 * i + 1] = '#';    }    s[0] = '*';    for(int i = 2; i < 2 * len + 1; i++)    {        if(p[id] + id > i) p[i] = min(p[id * 2 - i], p[id] - (i - id));        else p[i] = 1;        while(s[i + p[i]] == s[i - p[i]]) ++p[i];        if(p[i] + i > p[id] + id) id = i;        if(maxlen < p[i])        {            maxlen = p[i];            t = i + (p[i] - 1);            st = i - (p[i] - 1);        }    }    return;}int main(){    ios::sync_with_stdio(false);    cin.tie(0);    while(cin >> c >> s)    {        Turn(s);        Manacher(s);        if((maxlen - 1) >= 2)        {            cout << (st + 1)/2 - 1 << " " << (t - 1)/2 - 1 << endl;            for(int i = st + 1; i < t; i += 2)            {                cout << s[i];            }            cout << endl;        }        else        {            cout << "No solution!\n";        }    }}

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