Sample Input
2
abc
abaadada
abc
abaadada
Sample Output
Yes
No
No
判断是否能成为3个非空回文子串
manacher算法求出个点回文长度,在找出第一个和最后一个保存下来,再判断中间的
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MOD 3221225473
#define MAXN 20005
#define MIN 0
#define MAX 1000001 int n;
char d[MAXN];
char st[MAXN*2];
int p[MAXN*2],begi[MAXN*2],tail[MAXN*2];
int len;
void manacher()
{
int MaxId=0,id;
for(int i=0; i<len; i++)
{
if(MaxId>i)
p[i]=min(p[2*id-i],MaxId-i);
else
p[i]=1;
while(st[i+p[i]]==st[i-p[i]])
p[i]++; if(p[i]+i>MaxId)
{
id=i;
MaxId=p[i]+i;
}
p[i] -= 1;
}
} int main()
{
int T;
scanf("%d",&T);
while(T--)
{
getchar();
scanf("%s",d);
int l = strlen(d); len = 0;
st[len++] = '#';
for(int i = 0; i < l; i++)
{
st[len++] = d[i];
st[len++] = '#';
}
st[len] = 0;
manacher(); int flag = 0;
int pn =0 ,ln = 0;
for(int i = 1; i < len - 1; i++)
{
if(i - p[i] == 0) begi[pn++] = i;
if(i + p[i] == len-1) tail[ln++] = i;
}
for(int i = 0; i < pn; i++)
{
for(int j = ln - 1; j>=0; j--)
{
int s1 = begi[i] + p[begi[i]]+1 ,s2 = tail[j] - p[tail[j]]-1;
if(s1 > s2)
break;
int mid = (s1 + s2)/2;
if(p[mid] >= mid-s1)
{
flag = 1;
break;
}
}
if(flag )
break;
}
if(flag )
printf("Yes\n");
else
printf("No\n");
}
return 0;
}