hdu3294 Girls' research (manacher算法,最大回文子串)

2017年02月27日 11点热度 0人点赞 0条评论

Girls' research

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2164    Accepted Submission(s): 826


Problem Description
One day, sailormoon girls are so delighted that they intend to research about palindromic strings. Operation contains two steps:
First step: girls will write a long string (only contains lower case) on the paper. For example, "abcde", but 'a' inside is not the real 'a', that means if we define the 'b' is the real 'a', then we can infer that 'c' is the real 'b', 'd' is the real 'c' ……,
'a' is the real 'z'. According to this, string "abcde" changes to "bcdef".
Second step: girls will find out the longest palindromic string in the given string, the length of palindromic string must be equal or more than 2.
 


Input
Input contains multiple cases.
Each case contains two parts, a character and a string, they are separated by one space, the character representing the real 'a' is and the length of the string will not exceed 200000.All input must be lowercase.
If the length of string is len, it is marked from 0 to len-1.
 


Output
Please execute the operation following the two steps.
If you find one, output the start position and end position of palindromic string in a line, next line output the real palindromic string, or output "No solution!".
If there are several answers available, please choose the string which first appears.
 


Sample Input

b babd
a abcd
 


Sample Output

0 2
aza
No solution!
#include
#include
#include
using namespace std;
const int Max=400010;
int main(){
	char t,a[Max];
	int p[Max];
	while(~scanf(" %c%s",&t,a)){
		int len=strlen(a),id=0,mx=0,f;
		for(int i=len-1;i>=0;i--){
			a[i*2+2]=a[i]-(t-'a');
			if(a[i*2+2]<'a')
			a[i*2+2]+=26;
			a[i*2+1]='#';
		}
		a[0]='*';
		len=len*2+1;
		a[len+1]=0;
		for(int i=2;ii) p[i]=min(p[id]+id-i,p[id*2-i]);
			else p[i]=1;
			while(a[i-p[i]]==a[i+p[i]])
			p[i]++;
			if(p[i]+i>p[id]+id) id=i;
			if(p[i]>mx){
				mx=p[i];
				f=i;
			}
		}
		if(mx==2){
			puts("No solution!");
			continue;
		}
		printf("%d %dn",(f-mx+2)/2-1,(f+mx-2)/2-1);
		for(int i=f-mx+2;i
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update

纸上得来终觉浅, 绝知此事须躬行。