HDU - 4436sam裸题

Posted 2021-01-13 acjiumeng

题意:给你多个数字串,求本质不同的子串和(去掉前导零)
题解:建广义sam,刚开始一直想的是用l来计算,发现前导零对l的影响根本消不掉,所以不会做= =,原来应该是直接用一个新的数组表示到当前有多少个子串就好了

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 2012
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("c.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=100000+10,inf=0x3f3f3f3f;

char s[N];
struct SAM{
    int last,cnt;
    int ch[N<<1][10],fa[N<<1],l[N<<1],c[N<<1];
    int a[N<<1],num[N<<1],sum[N<<1];
    void init()
    {
        cnt=1;
    }
    void ins(int x)
    {
        if(ch[last][x])
        {
            int p=last,q=ch[last][x];
            if(l[q]==l[p]+1)last=q;
            else
            {
                int nq=++cnt;l[nq]=l[p]+1;
                memcpy(ch[nq],ch[q],sizeof ch[q]);
                fa[nq]=fa[q];fa[q]=last=nq;
                for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq;
            }
            return ;
        }
        int p=last,np=++cnt;last=np;l[np]=l[p]+1;
        for(;p&&!ch[p][x];p=fa[p])ch[p][x]=np;
        if(!p)fa[np]=1;
        else
        {
            int q=ch[p][x];
            if(l[q]==l[p]+1)fa[np]=q;
            else
            {
                int nq=++cnt;l[nq]=l[p]+1;
                memcpy(ch[nq],ch[q],sizeof ch[q]);
                fa[nq]=fa[q];fa[q]=fa[np]=nq;
                for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq;
            }
        }
    }
    void topo()
    {
        for(int i=1;i<=cnt;i++)c[l[i]]++;
        for(int i=1;i<=cnt;i++)c[i]+=c[i-1];
        for(int i=1;i<=cnt;i++)a[c[l[i]]--]=i;
    }
    void build()
    {
        last=1;
        int len=strlen(s+1);
        for(int i=1;i<=len;i++)ins(s[i]-'0');
    }
    void cal()
    {
        topo();
        num[1]=1;
        for(int i=1;i<=cnt;i++)
        {
            int p=a[i];
            for(int j=0;j<10;j++)
                if(ch[p][j])
                {
                    if(i==1&&j==0)continue;
                    num[ch[p][j]]+=num[p];
                    sum[ch[p][j]]=(sum[ch[p][j]]+(sum[p]*10+num[p]*j)%mod)%mod;
                }
        }
        int ans=0;
        for(int i=1;i<=cnt;i++)
            ans=(ans+sum[i])%mod;
        printf("%d
",ans);
        for(int i=0;i<=cnt;i++)
        {
            for(int j=0;j<10;j++)ch[i][j]=0;
            sum[i]=fa[i]=num[i]=a[i]=c[i]=l[i]=0;
        }
    }
}sam;
int main()
{
//    fin;
    int n;
    while(~scanf("%d",&n))
    {
        sam.init();
        for(int i=0;i<n;i++)
        {
            scanf("%s",s+1);
            sam.build();
        }
        sam.cal();
    }
    return 0;
}
/********************
3
12
012
0012
********************/