hdu5340—Three Palindromes—(Manacher算法)——回文子串
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Three Palindromes
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1948 Accepted Submission(s): 687
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
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
2
abc
abaadada
Sample Output
Yes
No
题意:给出一个字符串,问能否将字符串分为三段,并且每一段的是回文串。
思路:这题要用到manacher算法,不知道的可以看一下这篇博客:https://blog.csdn.net/dyx404514/article/details/42061017。
首先用manacher求出以每个点为中点的回文串的半径,然后,我们可以知道,要将字符串分为三段,那第一段一定包含第一个字符,第三段一定包含最后一个字符。然后判断第一个回文串和第三个之间剩下的字符是否构成回文串就行了。
所以,我们找到用manacher算法求出的所有回文串中,包含了第一个字符和和最后一个字符的有那些,然后两两配对,看看是否存在某一对,他们不重合,且之间剩下的字符也是回文串。
具体操作看代码:
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<string> 5 #include<cmath> 6 #include<algorithm> 7 #include<stack> 8 #include<queue> 9 #define eps 1e-7 10 #define ll long long 11 #define inf 0x3f3f3f3f 12 #define pi 3.141592653589793238462643383279 13 using namespace std; 14 const int maxn = 20007; 15 char a[maxn],s_new[maxn<<1]; 16 int s_len[maxn<<1]; 17 18 int Init() //manacher算法初始化(模板) 19 { 20 int len = strlen(a); 21 s_new[0] = ‘@‘; 22 int j = 1; 23 for(int i=0; i<len; ++i) //在每个字符左右插入‘#‘ 24 { 25 s_new[j++] = ‘#‘; 26 s_new[j++] = a[i]; 27 } 28 s_new[j++] = ‘#‘; 29 s_new[j] = ‘