SAM spoj SUBST1 New Distinct Substring

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

l[] 维护节点长度最大值
l[f[i]]=min[i]-1
在sam上的一个节点i上，有贡献是子串数量就是 max[i]-min[i]+1
因为只维护max，而min可以由f[i]的max+1得到
所以i贡献的答案是=max[i]-min[i]+1=l[i]-l[f[i]]

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<iostream>
using namespace std;
const int N=80005;
map<int,int>c[N];
int T,n,f[N],l[N],pre,cnt;
typedef long long ll;
char s[N];
ll ans;
inline void clr(){
    pre=cnt=1;ans=0;
    memset(f,0,sizeof(f));memset(l,0,sizeof(l));
    for(int i=0;i<N;++i)c[i].clear();
}
inline void ins(int x){
    int p=pre,np=++cnt;pre=np;
    l[np]=l[p]+1;
    for(;p&&!c[p].count(x);p=f[p])c[p][x]=np;
    if(!p)f[np]=1;
    else{
        int q=c[p][x];
        if(l[q]==l[p]+1)f[np]=q;
        else{
            int nq=++cnt;
            l[nq]=l[p]+1;
            c[nq]=c[q];f[nq]=f[q];
            f[q]=f[np]=nq;
            for(;c[p][x]==q;p=f[p]) c[p][x]=nq;
        }
    }
}
int main(){
    scanf("%d",&T);
    while(T--){clr();
        scanf("%s",s);
        n=strlen(s);
        for(int i=0;i<n;++i){
            ins(s[i]);
        }
        for(int i=1;i<=cnt;++i)ans+=(ll)(l[i]-l[f[i]]);
        cout<<ans<<endl;
    }
    return 0;
}