Description
The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing ?Robots Unlimited? has infiltrated into “U.S. Robotics”. ?U.S. Robots? security service would have already
started an undercover operation to establish the agent’s identity, but, fortunately, the letter describes communication channel the agent uses. He will publish articles containing stolen data to the “Solaris” almanac. Obviously, he will obfuscate the data,
so “Robots Unlimited” will have to use a special descrambler (“Robots Unlimited” part number NPRx8086, specifications are kept secret).
Having read the letter, the “U.S. Robots” president recalled having hired the “Robots Unlimited” ex-employee John Pupkin. President knows he can trust John, because John is still angry at being mistreated by “Robots Unlimited”.
Unfortunately, he was fired just before his team has finished work on the NPRx8086 design.
So, the president has assigned the task of agent’s message interception to John. At first, John felt rather embarrassed, because revealing the hidden message isn’t any easier than finding a needle in a haystack. However, after
he struggled the problem for a while, he remembered that the design of NPRx8086 was still incomplete. “Robots Unlimited” fired John when he was working on a specific module, the text direction detector. Nobody else could finish that module, so the descrambler
will choose the text scanning direction at random. To ensure the correct descrambling of the message by NPRx8086, agent must encode the information in such a way that the resulting secret message reads the same both forwards and backwards.
In addition, it is reasonable to assume that the agent will be sending a very long message, so John has simply to find the longest message satisfying the mentioned property.
Your task is to help John Pupkin by writing a program to find the secret message in the text of a given article. As NPRx8086 ignores white spaces and punctuation marks, John will remove them from the text before feeding it into
the program.
Input
The input consists of a single line, which contains a string of Latin alphabet letters (no other characters will appear in the string). String length will not exceed 1000 characters.
Output
The longest substring with mentioned property. If there are several such strings you should output the first of them.
Sample Input
| input | output |
|---|---|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA |
ArozaupalanalapuazorA |
Source
Problem Author: Eugene Krokhalev
Problem Source: IX Open Collegiate Programming Contest of the High School Pupils (13.03.2004)
还是论文里的题目
首先反转，拼接原串与反转串，然后求两者的追偿公共前缀，唯一注意的是如果有多组要输出最先出现那一组
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 2222
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int rank[N],height[N],s[N],a[N];
char str[N],str1[N],str2[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
int cmp(int *r,int a,int b,int k)
{
return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)//n要包含末尾添加的0
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[x[i]=r[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[x[i]]]=i;
p=1;
j=1;
for(; p<n; j*=2,m=p)
{
for(p=0,i=n-j; i<n; i++) y[p++]=i;
for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
for(i=0; i<n; i++) wv[i]=x[y[i]];
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[wv[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[wv[i]]]=y[i];
t=x;
x=y;
y=t;
x[sa[0]]=0;
for(p=1,i=1; i<n; i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
}
}
void getheight(int *r,int n)//n不保存最后的0
{
int i,j,k=0;
for(i=1; i<=n; i++) rank[sa[i]]=i;
for(i=0; i<n; i++)
{
if(k)
k--;
else
k=0;
j=sa[rank[i]-1];
while(r[i+k]==r[j+k])
k++;
height[rank[i]]=k;
}
}
int main()
{
int len,n=0,i,j,k;
W(~scanf("%s",str))
{
n = 0;
len = strlen(str);
UP(i,0,len-1)
s[n++] = str[i];
s[n++] = 200;
DOWN(i,len-1,0)
s[n++] = str[i];
s[n] = 0;
getsa(s,sa,n+1,300);
getheight(s,n);
int ans = 1,x=0;
UP(i,1,n-1)
{
int minn = min(sa[i],sa[i-1]);
int maxn = max(sa[i],sa[i-1]);
if(minn>=len || maxn<len) continue;//要分别在两个串中
if(minn+height[i]!=n-maxn) continue;//minn+height[i]是两者公共前缀的最后一个，而这个对应的位置是在后一串的n-（n-sa[k]），也就是n-maxn位置
if(height[i]>ans)
{
ans = height[i];
x = minn;
}
else if(height[i]==ans)
{
x = min(minn,x);
}
}
for(i=x; ans--; i++)
printf("%c",str[i]);
puts("");
}
}