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Front compression
Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1382 Accepted Submission(s): 517
Problem Description
Front compression is a type of delta encoding compression algorithm whereby common prefixes and their lengths are recorded so that they need not be duplicated. For example:
The size of the input is 43 bytes, while the size of the compressed output is
40<†·Ÿ"http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">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"t exceed 100,000. The second line contains a integer 1 ¡Ü N ¡Ü 100,000, which is the number of lines in the input. Each of the following N lines contains two
integers 0 ¡Ü A < B ¡Ü length(S), indicating that that line of the input is substring [A, B) of S.
Output
For each test case, output the sizes of the input and corresponding compressed output.
Sample Input
frcode
2
0 6
0 6
unitedstatesofamerica
3
0 6
0 12
0 21
myxophytamyxopodnabnabbednabbingnabit
6
0 9
9 16
16 19
19 25
25 32
32 37
Sample Output
14 12
42 31
43 40
Author
Zejun Wu (watashi)
Source
2013 Multi-University Training Contest 9
#include
#include
#include
#include
#include
using namespace std;
typedef long long int LL;
const int maxn=102100;
int sa[maxn],rank[maxn],rank2[maxn],h[maxn],c[maxn],*x,*y,ans[maxn];
char str[maxn];
bool cmp(int* r,int a,int b,int l,int n)
{
if(r[a]==r[b]&&a+l=0;i--) sa[--c[x[y[i]]]]=y[i];
}
void get_sa(char c[],int n,int sz=128)
{
x=rank,y=rank2;
for(int i=0;i=len) y[yid++]=sa[i]-len;
radix_sort(n,sz);
swap(x,y);
x[sa[0]]=yid=0;
for(int i=1;i=n) break;
}
for(int i=0;ir) swap(l,r);
///!!!!!if(l==r) return n-sa[l];
int a=l+1,b=r;
int k=Log[b-a+1];
return min(dp[a][k],dp[b-(1<>a>>b)
cout<<"lcp: "<
