hdu-3249-Girls' research -mancher

来源：互联网发布：云计算怎么赚钱编辑：程序博客网时间：2024/06/05 18:38

题目传送门：https://vjudge.net/problem/HDU-3294

就是manacher求一下回文串，同时记录一下起始位置和长度，输出。先进行字母的替换。

感觉，最不好推得是，根据变形后的字符串去找现在字符串的位置。

 l = (i - 1)/2 - (ans - 1)/2;            r = l + ans - 2;

ac代码：

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define N 200005#define ll long longusing namespace std;char a[N];char s[2*N];int p[2*N], len, l, r;int manacher(){    int pos = 0, maxr = 0, ans = 0;    for(int i = 2; i <= 2*len; i++)    {        p[i] = i < maxr ? min(p[2*pos-i], maxr -i) : 1;        while(s[i+p[i]] == s[i-p[i]])            p[i]++;        if(i + p[i] > maxr)            maxr = i + p[i], pos = i;        if(ans < p[i])        {            ans = p[i];            l = (i - 1)/2 - (ans - 1)/2;            r = l + ans - 2;        }    }    return ans - 1;}int main(){    char ch;    while(scanf("%c %s%*c", &ch, a+1)!=EOF)    {        len = strlen(a+1);        s[0] = '$';        s[1] = '#';        int c = ch;        for(int i = 1; i <= len; i++)        {            s[i*2+1] = '#';            int b = a[i];            s[i*2] = abs(b - c + 26) % 26 + 'a';        }        int ans = manacher();        if(ans <= 1)        {            printf("No solution!\n");        }        else        {            printf("%d %d\n", l, r);            for(int i = l+1; i <= r+1; i++)            {                int b = a[i];                printf("%c", abs(b - c + 26) % 26 + 'a');            }            printf("\n");        }    }    return 0;}

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