题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1297
1297. Palindrome
Time limit: 1.0 second
Memory limit: 64 MB
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 | output |
|---|---|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA |
ArozaupalanalapuazorA |
代码如下：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 2047
int wa[N], wb[N], wv[N], ws1[N];
int rank1[N]; //名次数组
int height[N]; //排名相邻的两个后缀的最长公共前缀
int sa[N]; //sa为后缀数组,n个后缀从小到大进行排序之后把排好序的后缀的开头位置
char s[N];
int a[N], n;
int dp[N][47];
int minn(int a,int b)
{
return a>b?b:a;
}
int cmp(int *r,int a,int b,int l)
{
return r[a]==r[b] && r[a+l]==r[b+l];
}
void get_sa(int *r, int *sa, int n, int m) //倍增算法
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0; i<m; i++) ws1[i]=0;
for(i=0; i<n; i++) ws1[x[i]=r[i]]++;
for(i=1; i<m; i++) ws1[i]+=ws1[i-1];
for(i=n-1; i>=0; i--) sa[--ws1[x[i]]]=i; //对长度为1的字符串排序
for(p=1,j=1; 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; //第二关键字排序结果
for(i=0; i<n; i++) wv[i]=x[y[i]];
for(i=0; i<m; i++) ws1[i]=0;
for(i=0; i<n; i++) ws1[wv[i]]++;
for(i=1; i<m; i++) ws1[i]+=ws1[i-1];
for(i=n-1; i>=0; i--) sa[--ws1[wv[i]]]=y[i]; //第一关键字排序
for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++; //更新rank数组
}
return;
}
void get_height(int *r, int *sa, int n) //求height数组
{
int i, j, k=0;
for(i=1; i<=n; i++) rank1[sa[i]]=i;
for(i=0; i<n; height[rank1[i++]]=k)
for(k?k--:0,j=sa[rank1[i]-1]; r[i+k]==r[j+k]; k++);
return;
}
void RMQ()
{
memset(dp,127,sizeof(dp));
for(int i = 1; i <= n*2+1; i++)
dp[i][0] = height[i];
for(int j = 1; (1<<j) <= 2*n+1; j++)
for(int i = 1; i+(1<<j)-1 <= 2*n+1; i++)
dp[i][j] = minn(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int lcp(int l,int r)//求最长公共前缀
{
int a = rank1[l], b = rank1[r];
if(a > b)
swap(a,b);
a++;
int t=(int)(log(double(b-a+1))/log(2.00));
return minn(dp[a][t],dp[b-(1<<t)+1][t]);
}
int main()
{
int res;
while(~scanf("%s",s))
{
int maxx = 0;
n = strlen(s);
for(int i = 0; i < n; i++)
a[i] = (int)s[i];
a[n] = 1;
for(int i = 0; i < n; i++)
a[i+n+1] = (int)s[n-i-1];
int LEN = 2*n+1;
a[LEN] = 0;
get_sa(a,sa,LEN+1,320);
get_height(a,sa,LEN);
RMQ();
int anslen = 0, idx = 0;
for(int i = 0; i < n; i++)
{
int t = lcp(i,LEN-i-1);
if(2*t-1 > anslen)//多算了一次中心字母
{
anslen = 2*t-1;
idx = i;
}
if(s[i] == s[i+1])
{
t = lcp(i+1,LEN-i-1);
if(2*t > anslen)
{
anslen = 2*t;
idx = i;
}
}
}
idx = idx-(anslen-1)/2;
for(int i = 0; i < anslen; i++)
printf("%c",s[idx+i]);
printf("\n");
}
return 0;
}