Codeforces Round #129 (Div. 1)E. Little Elephant and Strings
阿新 • • 發佈:2019-04-22
signed sss fail loop char s puts ack a* ons
題意:有n個串,詢問每個串有多少子串在n個串中出現了至少k次.
題解:sam,每個節點開一個set維護該節點的字符串有哪幾個串,啟發式合並set,然後在sam上走一遍該串,對於每個可行的串,所有的fail都是可行的直接加上len,不可行就往fail上跳.
被for(int i=0;s[i];i++)給騷死了,一直tle....
//#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 1000000007 #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 ull unsigned long long //#define base 1000000000000000000 #define fin freopen("a.txt","r",stdin) #define fout freopen("a.txt","w",stdout) #define fio ios::sync_with_stdio(false);cin.tie(0) inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;} inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;} 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 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 ull ba=233; const db eps=1e-5; const ll INF=0x3f3f3f3f3f3f3f3f; const int N=100000+10,maxn=1000000+10,inf=0x3f3f3f3f; int n,k; char s[N]; vector<char>v[N]; struct SAM{ int last,cnt; int ch[N<<1][26],fa[N<<1],l[N<<1]; int sz[N<<1],id[N<<1]; set<int>st[N<<1]; vi son[N<<1]; SAM(){cnt=1;} void ins(int c){ if(ch[last][c]) { int p=last,q=ch[last][c]; 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][c]==q;p=fa[p])ch[p][c]=nq; } return ; } int p=last,np=++cnt;last=np;l[np]=l[p]+1; for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np; if(!p)fa[np]=1; else { int q=ch[p][c]; if(l[p]+1==l[q])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][c]==q;p=fa[p])ch[p][c]=nq; } } } void build(int id) { last=1;int len=strlen(s); for(int i=0;i<len;i++)ins(s[i]-'a'),st[last].insert(id); } void dfs(int u) { for(int x:son[u]) { dfs(x); if(st[id[u]].size()<st[id[x]].size()) { for(int p:st[id[u]])st[id[x]].insert(p); st[id[u]].clear();id[u]=id[x]; } else { for(int p:st[id[x]])st[id[u]].insert(p); st[id[x]].clear(); } } sz[u]=st[id[u]].size(); } void cal() { for(int i=2;i<=cnt;i++)son[fa[i]].pb(i),id[i]=i; id[1]=1; dfs(1); for(int i=1;i<=n;i++) { int now=1;ll ans=0; for(int j=0;j<v[i].size();j++) { now=ch[now][v[i][j]-'a']; while(now!=1&&sz[now]<k)now=fa[now]; ans+=l[now]; } printf("%lld ",ans); } puts(""); } }sam; int main() { scanf("%d%d",&n,&k); for(int i=1;i<=n;i++) { scanf("%s",s); int n=strlen(s); for(int j=0;j<n;j++)v[i].pb(s[j]); sam.build(i); } sam.cal(); return 0; } /******************** ********************/
Codeforces Round #129 (Div. 1)E. Little Elephant and Strings