SPOJ 705 New Distinct Substrings
后缀数组。按照排序完的后缀一个一个统计。每一个后缀对答案做出的贡献为:n-SA[i]-height[i]。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-;
void File()
{
freopen("D:\\in.txt","r",stdin);
freopen("D:\\out.txt","w",stdout);
}
inline int read()
{
char c = getchar(); while(!isdigit(c)) c = getchar();
int x = ;
while(isdigit(c)) { x = x * + c - ''; c = getchar(); }
return x;
} const int maxn=+; int wa[maxn],wb[maxn],wv[maxn],WS[maxn];
int 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=wa,*y=wb,*t;
for(i=; i<m; i++) WS[i]=;
for(i=; i<n; i++) WS[x[i]=r[i]]++;
for(i=; i<m; i++) WS[i]+=WS[i-];
for(i=n-; i>=; i--) sa[--WS[x[i]]]=i;
for(j=,p=; p<n; j*=,m=p)
{
for(p=,i=n-j; i<n; i++) y[p++]=i;
for(i=; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
for(i=; i<n; i++) wv[i]=x[y[i]];
for(i=; i<m; i++) WS[i]=;
for(i=; i<n; i++) WS[wv[i]]++;
for(i=; i<m; i++) WS[i]+=WS[i-];
for(i=n-; i>=; i--) sa[--WS[wv[i]]]=y[i];
for(t=x,x=y,y=t,p=,x[sa[]]=,i=; i<n; i++)
x[sa[i]]=cmp(y,sa[i-],sa[i],j)?p-:p++;
}
return;
} int rank[maxn],height[maxn];
void calheight(int *r,int *sa,int n)
{
int i,j,k=;
for(i=; i<=n; i++) rank[sa[i]]=i;
for(i=; i<n; height[rank[i++]]=k)
for(k?k--:,j=sa[rank[i]-]; r[i+k]==r[j+k]; k++);
return;
} int T,n,a[maxn],SA[maxn];
char str[maxn]; int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%s",str); n=strlen(str);
for(int i=;i<n;i++) a[i]=(int)str[i];
a[n]=; da(a,SA,n+,); calheight(a,SA,n);
LL ans=;
for(int i=;i<=n;i++) ans=ans+n-SA[i]-height[i];
printf("%lld\n",ans);
}
return ;
}
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