后缀数组：SPOJ SUBST1

Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:
2
CCCCC
ABABA

Output:
5
9

　　题意：求一个字符串有多少不同子串
　　LCP的应用，会打后缀数组的模板后就十分简单了
　　后缀数组的理解：
　　http://www.cnblogs.com/staginner/archive/2012/02/02/2335600.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
const int maxn=50010;
char S[maxn];
int r[maxn],wa[maxn],wb[maxn],wv[maxn],ws[maxn],sa[maxn];

bool cmp(int *p,int i,int j,int l)
{return p[i]==p[j]&&p[i+l]==p[j+l];}

void DA(int n,int m)
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0;i<m;i++)
        ws[i]=0;
    for(i=0;i<n;i++)
        ++ws[x[i]=r[i]];
    for(i=1;i<m;i++)
        ws[i]+=ws[i-1];
    for(i=n-1;i>=0;i--)
        sa[--ws[x[i]]]=i;
        
    for(j=1,p=1;p<n;j<<=1,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++)
            ws[i]=0;    
        for(i=0;i<n;i++)
            ++ws[wv[i]];
        for(i=1;i<m;i++)            
            ws[i]+=ws[i-1];
        for(i=n-1;i>=0;i--)
            sa[--ws[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++;        
    }
}

int rank[maxn],lcp[maxn];
void LCP(int n)
{
    int i,j,k=0;
    for(i=1;i<=n;i++)
        rank[sa[i]]=i;
    for(i=0;i<n;lcp[rank[i++]]=k)
        for(k?k--:k,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);    
}

int main()
{
    int Q;
    scanf("%d",&Q);
    while(~scanf("%s",S))
    {
        int i,n;
        long long ans=0;
        for(i=0;S[i];i++)
        r[i]=S[i];
        DA(i+1,128);
        LCP(i);
        n=i;
        for(i=0;i<n;i++)
            ans+=n-i-lcp[rank[i]];
        printf("%d
",ans);    
    }
    return 0;
}