HDOJ 4691 Front compression 后缀数组

后缀数组求两子串间的最大公共前缀.

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. Here, every space and newline is also counted as 1 byte.
Given the input, each line of which is a substring of a long string, what are sizes of it and corresponding compressed output?

 

Input

There are multiple test cases. Process to the End of File.
The first line of each test case is a long string S made up of lowercase letters, whose length doesn't exceed 100,000. The second line contains a integer 1 ≤ N ≤ 100,000, which is the number of lines in the input. Each of the following N lines contains two integers 0 ≤ A < B ≤ 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 <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>

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<n&&b+l<n&&r[a+l]==r[b+l]) return true;
    return false;
}

void radix_sort(int n,int sz)
{
    for(int i=0;i<sz;i++) c[i]=0;
    for(int i=0;i<n;i++) c[x[y[i]]]++;
    for(int i=1;i<sz;i++) c[i]+=c[i-1];
    for(int i=n-1;i>=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<n;i++) x[i]=c[i],y[i]=i;
    radix_sort(n,sz);
    for(int len=1;len<n;len<<=1)
    {
        int yid=0;
        for(int i=n-len;i<n;i++) y[yid++]=i;
        for(int i=0;i<n;i++)
            if(sa[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;i++)
        {
            x[sa[i]]=cmp(y,sa[i],sa[i-1],len,n)?yid:++yid;
        }

        sz=yid+1;
        if(sz>=n) break;
    }
    for(int i=0;i<n;i++)
        rank[i]=x[i];
}

void get_h(char str[],int n)
{
    int k=0; h[0]=0x3f3f3f3f;
    for(int i=0;i<n;i++)
    {
        if(rank[i]==0) continue;
        k=max(k-1,0);
        int j=sa[rank[i]-1];
        while(i+k<n&&j+k<n&&str[i+k]==str[j+k]) k++;
        h[rank[i]]=k;
    }
}

int dp[maxn][20],Log[maxn];

void RMQ_init(int n)
{
    for(int i=0;i<n;i++) dp[i][0]=h[i];
    for(int i=1;i<=Log[n];i++)
    {
        for(int j=0;j+(1<<i)-1<n;j++)
        {
            dp[j][i]=min(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);
        }
    }
}

int lcp(int l,int r)
{
    l=rank[l];r=rank[r];
    if(l>r) 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<<k)+1][k]);
}

void init_Log()
{
    Log[0]=-1;
    for(int i=1;i<maxn;i++)
    {
        Log[i]=(i&(i-1))?Log[i-1]:Log[i-1]+1;
    }
}

void debug(int n)
{
    for(int i=0;i<n;i++)
    {
        cout<<sa[i]<<",";
    }
    cout<<endl;
    for(int j=0;j<n;j++)
    {
        cout<<h[j]<<",";
    }
    cout<<endl;
    for(int j=0;j<n;j++)
    {
        cout<<rank[j]<<",";
    }
    cout<<endl;
    int a,b;
    while(cin>>a>>b)
    cout<<"lcp: "<<lcp(a,b)<<endl;
}

int getwei(int x)
{
    if(x==0) return 1;
    int ret=0;
    while(x)
    {
        x/=10;
        ret++;
    }
    return ret;
}

int main()
{
    init_Log();
    while(scanf("%s",&str)!=EOF)
    {
        int n=strlen(str);
        get_sa(str,n);
        get_h(str,n);
        RMQ_init(n);

        //debug(n);

        int m;
        LL ans1=0,ans2=0;

        scanf("%d",&m);
        bool first=true;
        int lastL,lastR;
        while(m--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            ans1+=r-l+1LL;
            if(first==true)
            {
                ans2+=r-l+3;
                first=false;
            }
            else
            {
                if(lastL==l)
                {
                    int minR=min(lastR-1,r-1);
                    int com=minR-l+1;///公共长度
                    int diff=r-l+1-com;///不同的长度
                    int wei=getwei(com);

                    ans2+=diff+wei+1LL;
                    if(diff==0) ans2+=1LL;
                }
                else
                {
                    int com=lcp(lastL,l);///LCP
                    com=min(com,min(lastR-lastL,r-l));///公共长度
                    int diff=r-l+1-com;
                    int wei=getwei(com);
                    ans2+=diff+wei+1LL;
                    if(diff==0) ans2+=1LL;
                }
            }
            lastL=l; lastR=r;
        }
        cout<<ans1<<" "<<ans2<<endl;
    }
    return 0;
}