hdu4691(后缀数组)

算是后缀数组的入门题吧。 思路无比简单，要是直接套模板的话应该很容易秒掉。

关于后缀数组看高中神犇的论文就可以学会了

算法合集之《后缀数组——处理字符串的有力工具》

话说这题暴力是可以过了，但是我们在做多校的时候就是用暴力过的，当时还不知道什么是后缀数组。。。

靠着概念纯手敲了几个小时，把建SA，求height，和RMQ的ST算法都复习了一遍，这个东西要是每次都手敲的话真的会死人，尤其是倍增算法基数排序怎么排怎么别扭。自己写的倍增算法又太长，大牛的倍增算法总感觉敲的不顺。

贴个代码做留念。。。

Front compression

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1344    Accepted Submission(s): 500

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 <string.h>
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define N 100100

char g[N];
int r[N];
int sa[N];
int scnt[N];
int wa[N],wb[N],wv[N];
int mrank[N];
int h[N],th[N];
int dp[N][22];
int save[N];

int cmp(int gg[],int a,int b,int k)
{
	return gg[a]==gg[b] && gg[a+k]==gg[b+k];
}

void getsa(int str[],int sa[],int n,int m)
{
	int i,j,*x,*y,*t;
	x=wa; y=wb;
	memset(scnt,0,sizeof(scnt));
	for(i=0;i<n;i++)
		scnt[ x[i]=str[i] ]++;
	for(i=1;i<m;i++)
		scnt[i]+=scnt[i-1];
	for(i=0;i<n;i++)
		sa[ --scnt[ str[i] ] ]=i;

	for(int p=1,j=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] ];
		memset(scnt,0,sizeof(scnt));
		for(i=0;i<n;i++) scnt[ wv[i] ]++;
		for(i=1;i<m;i++) scnt[i]+=scnt[i-1];
		for(i=n-1;i>=0;i--) sa[ --scnt[ wv[i] ] ] = y[i];
		for(p=1,t=x,x=y,y=t,x[sa[0]]=0,i=1;i<n;i++)
			x[ sa[i] ] = cmp(y,sa[i],sa[i-1],j)?p-1:p++;
	}
}

void geth(int str[],int n)
{
	h[n-1]=0;
	int p=0;
	for(int i=0;i<n-1;i++)
	{
		int tmp=mrank[i];
		while( str[i+p] == str[ sa[tmp-1]+p ] ) p++;
		h[i]=p;
		p--;
		p=max(0,p);
	}
}

void buildst(int n)
{
	for(int i=1;i<=n;i++)
		dp[i][0] = th[i];
	for(int i=1;(1<<i)<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			if( j+(1<<(i-1)) >n ) dp[j][i]=dp[j][i-1];
			else dp[j][i]=min(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);
		}
	}
}

int cal(int x,int y)
{
	if(x>y) swap(x,y);
	x++;
	//然后就是求x到y的最小值
	int k=0;
	while( (1<<k)<=(y-x+1) ) k++;
	k--;
	return min(dp[x][k],dp[y-(1<<k)+1][k]);
}

int main()
{
	while(scanf("%s",g)!=EOF)
	{
		int len=strlen(g);
		for(int i=0;i<len;i++)
			r[i]=g[i];
		r[len++]=0;
		getsa(r,sa,len,300);
		for(int i=0;i<len;i++)
			mrank[ sa[i] ]=i;
		//for(int i=0;i<len;i++) printf("%s\n",g+sa[i]);
		geth(r,len);
		for(int i=0;i<len-1;i++)
			th[ mrank[i] ]= h[i];
		buildst(len-1);
		int m;
		long long ans1=0,ans2=0;
		scanf("%d",&m);
		int x,y,tmp;
		scanf("%d%d",&x,&y);
		ans1+=y-x;

		ans2+=y-x;

		save[1]=0;

		for(int i=2;i<=m;i++)
		{
			int tx,ty;
			scanf("%d%d",&tx,&ty);
			ans1+=ty-tx;
			int cnt;
			if(tx==x)
				cnt=len-1-tx;
			else
				cnt=cal(mrank[tx],mrank[x]);

			save[i]=min(ty-tx,min(y-x,cnt));
			ans2+=ty-tx-save[i];
			x=tx; y=ty;
		}
		ans1+=m;
		ans2+=m+m;
		for(int i=1;i<=m;i++)
		{
			ans2++;
			save[i]/=10;
			while(save[i])
			{
				ans2++;
				save[i]/=10;
			}
		}
		cout<<ans1<<" "<<ans2<<endl;
	}
    return 0;
}