Bestcoders 回文串 Manacher 算法

Three Palindromes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 871    Accepted Submission(s): 262

Problem Description

Can we divided a given string S into three nonempty palindromes?

 

Input

First line contains a single integer T≤20 which denotes the number of test cases.

For each test case , there is an single line contains a string S which only consist of lowercase English letters.1≤|s|≤20000

 

Output

For each case, output the "Yes" or "No" in a single line.

 

Sample Input

2abcabaadada

 

Sample Output

YesNo

 

Source

BestCoder Round #49 ($)

 

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#include <bits/stdc++.h>using namespace std;int t;int p[220100];char s[220100], c[221000];int len;void palindrome(){    len = strlen(s);    c[0] = '$';    for (int i = 0; i<len; i++)        c[i * 2 + 1] = '#', c[i * 2 + 2] = s[i];    c[len * 2 + 1] = '#';    len = len * 2 + 2;    c[len] = '\0';    int mx = 0, id = 0;    for (int i = 1; i<len; i++)    {        if (mx>i) p[i] = min(p[id * 2 - i], mx - i);        else p[i] = 1;        while (c[i + p[i]] == c[i - p[i]]) p[i]++;        if (i + p[i]>mx) mx = i + p[i], id = i;    }}int solve(){    for(int i=2; i<len; i++)    {        if(i - p[i] != 0) continue;        for(int j=i+p[i]; j<len; j++)        {            if(j-p[j]+1 <= i+p[i])            {                int x = j - ( i + p[i] );                int last = (len + j + x) >> 1;                if(last <= len - 2 && last + p[last] == len)                return 1;            }        }    }    return 0;}int main(){    scanf("%d", &t);    while (t--)    {        scanf("%s", s);        palindrome();        if (solve()) printf("Yes\n");        else printf("No\n");    }    return 0;}