1040 Longest Symmetric String （25 分）

1040 Longest Symmetric String （25 分）

Given a string, you are supposed to output the length of the longest symmetric sub-string. For example, given Is PAT&TAP symmetric?, the longest symmetric sub-string is s PAT&TAP s, hence you must output 11.

Input Specification:

Each input file contains one test case which gives a non-empty string of length no more than 1000.

Output Specification:

For each test case, simply print the maximum length in a line.

Sample Input:

Is PAT&TAP symmetric?

Sample Output:

马拉车模板题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1010;
char beginn[maxn],endd[maxn];
int len[maxn];
int  tralation()
{int gg=strlen(beginn);int noww=0;endd[0]='&';for(int i=0;i<gg;i++){endd[++noww]='#';endd[++noww]=beginn[i];}endd[++noww]='#';return noww;
}
int Manacher(int total)
{int maxx = 0;int mx = 0,id = 0; //id表示的最大回文的中心点是哪一个 而mx表示的是最大回文的中心点的最远的边界是哪一个 for(int i =1;i<=total;i++){if(i<mx) //如果此刻 i的点比mx还要小的话说明 在mx-i处这边是回文 然后在比较一下 len[i] = min(mx - i,len[2*id-i]); //因为2*id - i 和 i 他们是相对于id对称的 所以说吧 就是要比较mx-i和len[2*id-i] ;else len[i] = 1; //如果此刻i的点比边界还要大的话 那就需要从一开始加了 while(endd[i+len[i]]==endd[i-len[i]])len[i]++;if(i+len[i]>mx){mx = i+len[i];id = i; }   maxx = max(maxx,len[i]);} return maxx-1;
}
int main()
{scanf("%[^\n]",beginn);int total = tralation();int ans = Manacher(total);printf("%d\n",ans);return 0;
}

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