Front compression(后缀数组+rmq)-优快云博客

本文链接：https://blog.youkuaiyun.com/spark__s/article/details/52949350

Front compression

Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 2109 Accepted Submission(s): 731

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

Author

Zejun Wu (watashi)

Source

2013 Multi-University Training Contest 9

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后缀数组应用之一：熟悉掌握后缀数组中各个数组的意义。对于求区间最小，我们用rmq算法即可。存个模板吧。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
const int maxn = 100100;
#define N 100100 
int maxl[N][30], minl[N][30];  
int _rank[maxn],wb[maxn],wv[maxn],wss[maxn];
int height[maxn]; 
int n;
void S_table()  
{  
    int l = int(log((double)n)/log(2.0));  
    for (int j=1;j<=l;j++)  
    {  
        for (int i=1; i + (1 << (j-1) ) - 1 <=n;++i)  
        {  
            maxl[i][j] = max(maxl[i][j-1], maxl[i + (1 << (j-1) )][j-1]);  
            minl[i][j] = min(minl[i][j-1], minl[i + (1 << (j-1) )][j-1]);  
        }  
    }  
}   
int rmq(int l, int r)  
{  
    int k = int(log((double)(r-l+1))/log(2.0));  
    int a1 = max(maxl[l][k], maxl[r - (1<<k) + 1][k]);  
    int a2 = min(minl[l][k], minl[r - (1<<k) + 1][k]);  
   // printf("Max: %d Min: %d\n", a1, a2);  
	return a2;
}   
void calheight(int *r,int *sa,int n)  
{  
	int i,j,k=0;  
	for(i=1;i<=n;i++) _rank[sa[i]]=i;  
	for(i=0;i<n;height[_rank[i++]]=k)  
		for(k?k--:0,j=sa[_rank[i]-1];r[i+k]==r[j+k];k++);  
	 return;  
}  
bool cmp(int *r,int a,int b,int l)
{
    return r[a]==r[b] && r[a+l]==r[b+l];
}
void da(int *r,int *sa,int n,int m)
{
    int i,j,p,*x=_rank,*y=wb,*t;
    for(i=0;i<m;i++) wss[i]=0;
    for(i=0;i<n;i++) wss[x[i]=r[i]]++;
    for(i=1;i<m;i++) wss[i]+=wss[i-1];
    for(i=n-1;i>=0;i--)
       sa[--wss[x[i]]]=i;

    for(j=1,p=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;
        for(i=0;i<n;i++) wv[i]=x[y[i]];
        for(i=0;i<m;i++) wss[i]=0;
        for(i=0;i<n;i++) wss[wv[i]]++;
        for(i=1;i<m;i++) wss[i]+=wss[i-1];
        for(i=n-1;i>=0;i--) sa[--wss[wv[i]]]=y[i];
        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++;
    }
    return;
}
long long wei(int s)
{
	long long ans=0;
	if(s==0)
		ans++;
	int n=s;
	while(n>0)
	{
		n=n/10;
		ans++;
	}
	return ans;
}
char s[maxn];
int l;
int r[maxn],sa[maxn];
int main()
{

    while(cin>>s)
    {
	    l = strlen(s);   l++;
        for(int i=0; i<l-1; i++) r[i] = s[i]-'a'+1;
        r[l-1] = 0;
        da(r,sa,l,27);
		calheight(r,sa,l-1); 
	/*	  for(i=0; i<l-1; i++)  // rank[i] : suffix(i)排第几
           printf("rank[%d] =  %d\n",i,_rank[i]);
        printf("\n");
        for(i=0; i<l; i++)   // sa[i] : 排在第i个的是谁
           printf("sa[%d] =  %d\n",i,sa[i]);
		system("pause");*/
		for(int i=1;i<l;i++)
		{
			maxl[i][0]=minl[i][0]=height[i];
		}
		n=l-1;
		S_table();
		int query;
		scanf("%d",&query);
		int left,right;
		scanf("%d%d",&left,&right);
		long long ans=2+(right-left)+1;
		long long ans1=right-left;
		//cout<<ans<<endl;
		int max_=right-left;
		int pre=left;
		for(int i=1;i<query;i++)
		{
			scanf("%d%d",&left,&right);
			ans1=ans1+(right-left);
			max_=min(max_,right-left);
			int nowsuff=left;
			int rr=max(_rank[pre],_rank[nowsuff]);
			int ll=min(_rank[pre],_rank[nowsuff]);
			int aa;
			if(pre==nowsuff)
			{
				aa=max_;
			}
			else
				aa=rmq(ll+1,rr);
			//cout<<height[3]<<' '<<height[4]<<endl;
			if(aa>max_)
				aa=max_;
			//cout<<aa<<"yws"<<endl;
			int add=wei(aa);
			ans=ans+add+1+(right-left-aa)+1;
			max_=right-left;
			pre=left;
			//cout<<ans<<endl;
		}
       cout<<(long long)ans1+query<<" "<<ans<<endl;	   
    }
    return 0;
}