hdu5558 后缀数组

Alice's Classified Message

Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 312 Accepted Submission(s): 122


[align=left]Problem Description[/align]
Alice wants to send a classified message to Bob. She tries to encrypt the message with her original encryption method. The message is a string S, which consists of N lowercase letters.

S[a…b] means a substring of S ranging from S[a] to S (0≤a≤b<N). If the first i letters have been encrypted, Alice will try to find a magic string P. Assuming P has K letters, P is the longest string which satisfies P=S[T...T+K−1] (0≤T<i，T+K≤N) and P=S[i…i+K−1](i+K≤N). In other words, P is a substring of S, of which starting address is within [0...i−1], and P is also a prefix of S[i...N−1]. If P exists, Alice will append integer K and T to ciphertext. If T is not unique, Alice would select the minimal one. And then i is incremented by K. If P does not exist, Alice will append -1 and the ASCII code of letter S[i] to ciphertext, and then increment i by 1.

Obviously the first letter cannot be encrypted. That is to say, P does not exist when i=0. So the first integer of ciphertext must be -1, and the second integer is the ASCII code of S[0].

When i=N, all letters are encrypted, and Alice gets the final ciphertext, which consists of many pairs of integers. Please help Alice to implement this method.

[align=left]Input[/align]
The first line of input contains an integer T, which represents the number of test cases (T≤50). Each test case contains a line of string, which has no more than 100000 lowercase letters. It is guaranteed that the total length of the strings is not greater than 2×106.

[align=left]Output[/align]
For each test case, output a single line consisting of “[b]Case #X:” first. X is the test case number starting from 1. Output the ciphertext in the following lines. Each line contains two integers separated by a single space.

[align=left]Sample Input[/align]

2

aaaaaa

aaaaabbbbbaaabbc

[align=left]Sample Output[/align]

Case #1:

-1 97

5 0

Case #2:

-1 97

4 0

-1 98

4 5

5 2

-1 99

/*
hdu5558  后缀数组

从[1,n]对于每个i，求suff[j](j < i)与suff[i]的最长公共前缀，
如果有多个，取最小的那个

我们可以通过后缀数组先求出，如果i-1和i，i和i+1都有公共前缀，
那么i-1和i+1也有公共前缀，所以可以先处理出每个i的左右界限。然后对于i左右扫描一下即可

然后枚举i，从pre[i]-nex[i]找到合适的结果即可

hhh-2016-03-10 18:17:04
*/
#include <algorithm>
#include <cmath>
#include <queue>
#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <vector>
#include <functional>
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std;
typedef long long ll;
const int maxn = 150050;

int t1[maxn],t2[maxn],c[maxn];
bool cmp(int *r,int a,int b,int l)
{
return r[a]==r[b] &&r[l+a] == r[l+b];
}

void get_sa(int str[],int sa[],int Rank[],int height[],int n,int m)
{
n++;
int p,*x=t1,*y=t2;
for(int i = 0; i < m; i++) c[i] = 0;
for(int i = 0; i < n; i++) c[x[i] = str[i]]++;
for(int i = 1; i < m; i++) c[i] += c[i-1];
for(int i = n-1; i>=0; i--) sa[--c[x[i]]] = i;
for(int j = 1; j <= n; j <<= 1)
{
p = 0;
for(int i = n-j; i < n; i++) y[p++] = i;
for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i]-j;
for(int i = 0; i < m; i++) c[i] = 0;
for(int i = 0; i < n; i++) c[x[y[i]]]++ ;
for(int i = 1; i < m; i++) c[i] += c[i-1];
for(int i = n-1; i >= 0; i--)  sa[--c[x[y[i]]]] = y[i];

swap(x,y);
p = 1;
x[sa[0]] = 0;
for(int i = 1; i < n; i++)
x[sa[i]] = cmp(y,sa[i-1],sa[i],j)? p-1:p++;
if(p >= n) break;
m = p;
}
int k = 0;
n--;
for(int i = 0; i <= n; i++)
Rank[sa[i]] = i;
for(int i = 0; i < n; i++)
{
if(k) k--;
int j = sa[Rank[i]-1];
while(str[i+k] == str[j+k]) k++;
height[Rank[i]] = k;
}
}

int pre[maxn],nex[maxn];
int Rank[maxn],height[maxn];
int sa[maxn],str[maxn];
char a[maxn];
int len;

int main()
{
int T,cas = 1;
scanf("%d",&T);
while(T--)
{
scanf("%s",a);
int len = 0;
for(int i =0;a[i] != '\0'; i++)
{
str[len++] = a[i]-'a'+1;
}
str[len] = 0;
get_sa(str,sa,Rank,height,len,30);

for(int i = 1; i <= len; i++)
{
if(height[i] == 0)
pre[i] = i;
else
pre[i] = pre[i-1];
}

for(int i = len; i >= 1; i--)
{
if(height[i+1] == 0 || i == len) nex[i] = i;
else nex[i] = nex[i+1];
}

int i = 0;
printf("Case #%d:\n",cas++);
while(i < len)
{
int now = Rank[i];   //i的排名
int k = 0,t = i;
int mi = height[now];
for(int j = now-1; j >= pre[now]; j--)
{
mi = min(mi,height[j+1]);
if(mi < k)
break;
if(sa[j] < i)
{
if(mi > k || (mi==k && sa[j] < t))
{
k = mi;
t = sa[j];
}
}
}
if(now+1 <= nex[now]) mi = height[now+1];
for(int j = now+1; j <= nex[now]; j++)
{
mi = min(mi,height[j]);
if(mi < k)
break;
if(sa[j] < i)
{
if(mi > k || (mi==k && sa[j] < t))
{
t = sa[j];
k = mi;
}
}
}

if(k == 0) printf("-1 %d\n",a[i]);
else printf("%d %d\n",k,t);
if(k) i+=k;
else i++;
}
}
return 0;
}