Ural1297 Palindrome（后缀数组）

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　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

【思路】

最长回文子串。

将字符串反向拼接在后，中间用一个没有出现的字符隔开，则问题转化为求新字符串两个特定后缀的lcp，枚举对称点i，对称数为奇的情况对应求lcp(i,n-i)，对称数为偶的情况对应求lcp(i,n-i-1)。

如图所示：



两个后缀的lcp可以用Sparse Table算法（倍增）在O(nlogn)时间内求解。

【代码】

1 #include<cstdio>
2 #include<cstring>
3 #include<iostream>
4 #define FOR(a,b,c) for(int a=(b);a<=(c);a++)
5 using namespace std;
6
7 const int maxn = 3000+10;
8 const int maxd = 22;
9
10 int s[maxn];
11 int sa[maxn],c[maxn],t[maxn],t2[maxn];
12
13 void build_sa(int m,int n) {
14     int i,*x=t,*y=t2;
15     for(i=0;i<m;i++) c[i]=0;
16     for(i=0;i<n;i++) c[x[i]=s[i]]++;
17     for(i=1;i<m;i++) c[i]+=c[i-1];
18     for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;
19
20     for(int k=1;k<=n;k<<=1) {
21         int p=0;
22         for(i=n-k;i<n;i++) y[p++]=i;
23         for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
24
25         for(i=0;i<m;i++) c[i]=0;
26         for(i=0;i<n;i++) c[x[y[i]]]++;
27         for(i=0;i<m;i++) c[i]+=c[i-1];
28         for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];
29
30         swap(x,y);
31         p=1; x[sa[0]]=0;
32         for(i=1;i<n;i++)
33             x[sa[i]]=y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k]?p-1:p++;
34         if(p>=n) break;
35         m=p;
36     }
37 }
38 int rank[maxn],height[maxn];
39 void getHeight(int n) {
40     int i,j,k=0;
41     for(i=0;i<=n;i++) rank[sa[i]]=i;
42     for(i=0;i<n;i++) {
43         if(k) k--;
44         j=sa[rank[i]-1];
45         while(s[j+k]==s[i+k]) k++;
46         height[rank[i]]=k;
47     }
48 }
49 int A[maxn][maxd];
50 void RMQ_init(int n) {
51     for(int i=1;i<=n;i++) A[i-1][0]=height[i];
52     for(int k=1;(1<<k)<=n;k++)
53         for(int i=0;(i+(1<<k))<=n;i++)
54             A[i][k]=min(A[i][k-1],A[i+(1<<(k-1))][k-1]);
55 }
56 int query(int l,int r) {
57     int k=0;
58     while(1<<(k+1)<=(r-l+1)) k++;
59     return min(A[l][k],A[r-(1<<k)+1][k]);
60 }
61 int lcp(int a,int b) {
62     int l=rank[a],r=rank[b];
63     if(r<l) swap(l,r); l--,r--;
64     if(r<0) return 0;
65     return query(l+1,r);    //l+1
66 }
67
68 int n;
69 char expr[maxn];
70
71 int main() {
72     while(scanf("%s",expr)==1) {
73         int len=strlen(expr),n=2*len+1;
74         for(int i=0;i<len;i++)s[i]=expr[i];
75         s[len]=1;
76         for(int i=0;i<len;i++)s[i+len+1]=expr[len-1-i];
77         s
=0;
78
79         build_sa('z'+1,n+1);
80         getHeight(n);
81         RMQ_init(n);
82         int ans=0,front,tmp;
83         for(int i=0;i<n;i++) {
84             tmp=lcp(i,n-i-1);
85             if(2*tmp-1>ans) {  //对称个数为奇数
86                 ans=2*tmp-1;
87                 front=i-tmp+1;
88             }
89             tmp=lcp(i,n-i);
90             if(2*tmp>ans) {  //对称个数为偶数
91                 ans=2*tmp;
92                 front=i-tmp;
93             }
94         }
95         expr[front+ans]='\0';
96         printf("%s\n",expr+front);
97     }
98     return 0;
99 }