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

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

2 abc abaadada

 

 

Sample Output

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;
}