I - O‘My!
Note: this is a harder version of Mirrored string I.
The gorillas have recently discovered that the image on the surface of the water is actually a reflection of themselves. So, the next thing for them to discover is mirrored strings.
A mirrored string is a palindrome string that will not change if you view it on a mirror.
Examples of mirrored strings are "MOM", "IOI" or "HUH". Therefore, mirrored strings must contain only mirrored letters {A, H, I, M, O, T, U, V, W, X, Y} and be a palindrome.
e.g. IWWI, MHHM are mirrored strings, while IWIW, TFC are not.
A palindrome is a string that is read the same forwards and backwards.
Given a string S of length N, help the gorillas by printing the length of the longest mirrored substring that can be made from string S.
A substring is a (possibly empty) string of characters that is contained in another string S. e.g. "Hell" is a substring of "Hello".
Input
The first line of input is T – the number of test cases.
Each test case contains a non-empty string S of maximum length 1000. The string contains only uppercase English letters.
Output
For each test case, output on a line a single integer -
the length of the longest mirrored substring that can be made from
string S.
Example
Input
3IOIKIOOIROQWOWMAN
Output
413思路：将镜面字母拆分出来，一个一个求最长回文串，差点因为数组重复而runtime error 没过，还好发现了
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 1000009 ;
char a[10009] , b[20009] ;
int p[20009] ;
int vis[27] ;
char c[20] = {‘A‘, ‘H‘, ‘I‘, ‘M‘, ‘O‘, ‘T‘, ‘U‘, ‘V‘, ‘W‘, ‘X‘, ‘Y‘};
char d[10009] ;
int cal(char *a , int lena)
{
b[0] = ‘$‘;
b[1] = ‘#‘;
int l = 2 ;
for(int i = 0 ; i < lena ; i++)
{
b[l++] = a[i];
b[l++] = ‘#‘;
}
b[l] = ‘\0‘;
// cout << b << endl ;
int mx = -1 ;
int mid , i ;
int len = 0 ;
for(int i = 1 ; i < l ; i++)
{
if(mx > i)
{
p[i] = min(p[mid * 2 - i] , mx - i);
}
else
{
p[i] = 1 ;
}
while(b[i-p[i]] == b[i+p[i]])
{
p[i]++;
}
if(mx < p[i] + i)
{
mid = i ;
mx = p[i] + i ;
}
len = max(len , p[i] - 1);
}
return len ;
}
int main()
{
for(int i = 0 ; i < 11 ; i++)
{
int x = c[i] - ‘A‘;
vis[x] = 1 ;
}
int n ;
scanf("%d" , &n);
while(n--)
{
scanf("%s" , a);
int lena = strlen(a);
int ans = 0 ;
int l = 0 ;
for(int i = 0 ; i < lena ; i++)
{
if(vis[a[i] - ‘A‘])
{
d[l++] = a[i];
}
if(!vis[a[i] - ‘A‘] || i == lena - 1)
{
if(l >= 1)
ans = max(ans ,cal(d , l));
d[l] = 0 ;
l = 0 ;
continue ;
}
}
printf("%d\n" , ans);
}
return 0 ;
}
原文地址：https://www.cnblogs.com/nonames/p/11291334.html