hdu 3294 Girls' research

http://acm.hdu.edu.cn/showproblem.php?pid=3294

Girls' research

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2443 Accepted Submission(s): 944


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

b babd

a abcd

[align=left]Sample Output[/align]

0 2
aza
No solution!
求最长回文字串的左右所处坐标，并输出该回文字串，当然需要预处理一下，刚开始所接受字符为初始化字符a
manacher算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
int p[2000006];
char a[2000006],b;
int main()
{
while(cin>>b>>a)
{
memset(p,0,sizeof(p));
int len=strlen(a),id=0,maxlen=0,cnt=0;
for(int i=len;i>=0;--i)
{
a[(i<<1)+1]='#';
a[(i<<1)+2]=a[i];
}
for(int i=2;i<2*len-1;i++)
{
if(p[id]+id>i) p[i]=min(p[id*2-i],p[id]+id-i);
else p[i]=1;
while(a[i-p[i]]==a[i+p[i]]) ++p[i];
if(id+p[id]<i+p[i])id=i;
if(maxlen<p[i]){
maxlen=p[i];
cnt=i;
}
}
if(maxlen-1<2)puts("No solution!");
else
{
cout<<(cnt-p[cnt]+1)/2<<' '<<(cnt+p[cnt]-3)/2<<endl;
int k=b-'a';
for(int i=cnt-p[cnt]+2;i<=cnt+p[cnt]-2;i+=2)
{
cout<<(char)((a[i]-'a'-k+26)%26+'a');
}
cout<<endl;
}
}
return 0;
}

暴力求解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
char a[1000006],b;
int main()
{
while(scanf("%c %s",&b,a)!=EOF)
{
getchar();
int len=strlen(a);
int maxlen=0,k=b-'a',l=0,r=0;
for(int i=0;i<len;i++)
{
for(int j=0;i-j>=0 && i+j<len;j++)
{
if(a[i-j]!=a[i+j])break;
if(maxlen<j*2+1)
{
maxlen=j*2+1;
l=i-j;
r=i+j;
}
}
for(int j=0;i-j>=0 && i+j+1<len;j++)
{
if(a[i-j]!=a[i+j+1])break;
if(maxlen<j*2+2)
{
maxlen=j*2+2;
l=i-j;
r=i+j+1;
}
}
}
if(maxlen<2)printf("No solution!\n");
else
{
cout<<l<<' '<<r<<endl;
for(int i=l;i<=r;i++)
cout<<(char)((a[i]-'a'-k+26)%26+'a');
cout<<endl;
}
}
return 0;
}