# hdu 6194 沈阳网络赛--string string string(后缀数组)

Problem Description
Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.
Given a string s, we define a substring that happens exactly

Input
The first line contains an integer

Output
For each test case, print the number of the important substrings in a line.

Sample Input
2
2
abcabc
3
abcabcabcabc

Sample Output
6
9

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
const int N=1e5+5;
char s[N];
int k;
int wa[N],wb[N],wv[N],wss[N];
int sa[N],ran[N],height[N];
int f[N][20];

int cmp(int *r,int a,int b,int l)
{
return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(char *r,int *sa,int n,int m)
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0; i<m; i++) wss[i]=0;
for(i=0; i<n; i++) wss[x[i]=(int)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;
}
void callheight(char *r,int *sa,int n)
{
int i,j,k=0;
for(i=1;i<=n;i++)
ran[sa[i]]=i;
for(i=0;i<n;height[ran[i++]]=k)
for(k?k--:0,j=sa[ran[i]-1];r[i+k]==r[j+k];k++);
return ;
}
void init(int len)
{
for(int i=1;i<=len;i++) f[i][0]=height[i];
for(int s=1;(1<<s)<=len;s++)
{
int tmp=(1<<s);
for(int i=1;i+tmp-1<=len;i++)
{
f[i][s]=min(f[i][s-1],f[i+tmp/2][s-1]);
}
}
}
int cal(int l,int r)
{
int len=log2(r-l+1);
int ans=min(f[l][len],f[r-(1<<len)+1][len]);
return ans;
}
int main()
{
int T; cin>>T;
while(T--)
{
scanf("%d%s",&k,s);
int len=strlen(s);
da(s,sa,len+1,130);
callheight(s,sa,len);
init(len);
int ans=0;
for(int i=1;i+k-1<=len;i++)
{
int j=i+k-1;
int tmp=height[i];
if(j+1<=len) tmp=max(tmp,height[j+1]);
int x;
if(k!=1) { x=cal(i+1,j); }
else x=len-sa[i];
ans+=max(0,x-tmp);
}
printf("%d\n",ans);
}
return 0;
}

posted @ 2017-09-11 18:58 茶飘香~ 阅读(...) 评论(...) 编辑 收藏