Hdu 5340 Three Palindromes 最大回文串 Manacher

Three Palindromes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 80    Accepted Submission(s): 21


[align=left]Problem Description[/align]
Can we divided a given string S into three nonempty palindromes?

 

[align=left]Input[/align]
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

 

[align=left]Output[/align]
For each case, output the "Yes" or "No" in a single line.

 

[align=left]Sample Input[/align]

2
abc
abaadada

 

[align=left]Sample Output[/align]

Yes
No

 题意，是给出一个字符串，能否分成三个非空回文串。
我们可以发现 第一个和第三个串，一定是最大回文串的某个串,Manacher 求出最大回文串的长度，枚举第一个和最后一个，中间直接判断，中点的最大回文串是否包括了就可以了。复杂度为o(n * n).
具体manacher算法参见 Manacher算法

#define N 110050
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int T,n,a,pri
,ans,len,sn = 0,top
,tail
,pn,ln;
bool dp
[4];
pii seg
;
bool Manacher(char str[],int len){
char tstr[N+N];
int p[N + N],l2 =0,mi;
tstr[l2++] = '#';
for(int i =0;i<len;i++){
tstr[l2++] = str[i];
tstr[l2++] = '#';
}
p[0] = 0;mi = 0;
for(int i = 1;i<l2;i++){
int mi2 = mi + mi - i;
if(mi + p[mi] >= i) p[i] = min(mi2 - (mi - p[mi]),p[mi2]);
else p[i] = 0;
if(p[i] == 0 ||  mi2 - p[mi2] == mi - p[mi]){
int maxx = p[i]+1;
while(i- maxx >= 0 && i+maxx < l2 && tstr[i-maxx] == tstr[i+maxx]){
maxx++;
}
p[i] = maxx - 1;
}
if(p[i] + i > p[mi] + mi) mi = i;
}
int ans = -1;sn = 0;pn = ln = 0;
for(int i = 1;i < l2 - 1;i++){
if(i - p[i] == 0) top[pn++]  = i;
if(i + p[i] == l2 - 1) tail[ln++] = i;
}
for(int i = 0;i < pn;i++){
for(int j = ln - 1;j>=0;j--){
int s1 = top[i] + p[top[i]] + 1,s2 = tail[j] - p[tail[j]] - 1;
if(s1 > s2 )
break;
int mid = (s1 + s2)/2;
if(p[mid] >= mid - s1) return true;
}
}
return false;
//printf("%d\n",ans);
}
char str
;
int main()
{
while(S(T)!=EOF)
{
while(T--){
SS(str);
len = strlen(str);
if(Manacher(str,len))
printf("Yes\n");
else
printf("No\n");
}
}
return 0;
}

[align=left]Source[/align]
BestCoder Round #49 ($)
 

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