HDU 5340——Three Palindromes——————【manacher处理回文串】

Three Palindromes

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1244 Accepted Submission(s): 415


[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

题目大意：问是否可以找出三段回文串。

题解：


没有进行暴力压位，时间接近超时。但是很侥幸过了。

#include<bits/stdc++.h>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
const int maxn=20200;
int pre[maxn*2],suf[maxn*2];
int p[maxn*2];
char str[maxn],trans[maxn*2];
int Transform(){
//  memset(p,0,sizeof(p));
memset(pre,0,sizeof(pre));
memset(suf,0,sizeof(suf));
int len=strlen(str);
trans[0]='$';
for(int i=1;i<=2*len;i+=2){
trans[i]='#';
trans[i+1]=str[i/2];
}
trans[2*len+1]='#';
trans[2*len+2]='@';
return 2*len+1;
}
int manacher(){
int len=Transform();
int mx=0,pos=0;
for(int i=1;i<=len;i++){
if(i<mx){
p[i]=min(p[2*pos-i],mx-i);
}else{
p[i]=1;
}
for(;trans[i+p[i]]==trans[i-p[i]];p[i]++);
if(mx<i+p[i]){
mx=i+p[i];
pos=i;
}
}
return len;
}
int main(){
int t;
scanf("%d",&t);
while(t--){
scanf("%s",str);
int lens=strlen(str);
if(lens<3){
printf("No\n");continue;
}else if(lens==3){
printf("Yes\n");continue;
}else{
int len= manacher();
for(int i=2;i<len;i++){
if(p[i]==i){
pre[i+p[i]-1]=1;
}
if(p[i]==len-i+1){
suf[i-p[i]+1]=1;
}
}
int flag=0;
for(int i=2;i<len&&(!flag);i++){
for(int j=1;j<p[i]&&(!flag);j++){
if(pre[i-j]&suf[i+j]){
flag=1;
printf("Yes\n");
}
}
}
if(!flag){
printf("No\n");
}
}
}
return 0;
}