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  • CF204E Little Elephant and Strings [SAM,set启发式合并,子树数颜色]

    如果颜色数 (geq) k 就可以 (cnt_u += len_u - len_{fa_u}) 也就是这个点包含的字符串数量。
    然后 (cnt_u) 表示 u 节点对应的字符串所包含的子串的贡献,这样就可以算完了。
    就很简单了。

    // clang-format off
    // powered by c++11
    // by Isaunoya
    #include<bits/stdc++.h>
    #define rep(i,x,y) for(register int i=(x);i<=(y);++i)
    #define Rep(i,x,y) for(register int i=(x);i>=(y);--i)
    using namespace std;using db=double;using ll=long long;
    using uint=unsigned int;using ull=unsigned long long;
    using pii=pair<int,int>;
    #define Tp template
    #define fir first
    #define sec second
    Tp<class T>void cmax(T&x,const T&y){if(x<y)x=y;}Tp<class T>void cmin(T&x,const T&y){if(x>y)x=y;}
    #define all(v) v.begin(),v.end()
    #define sz(v) ((int)v.size())
    #define pb emplace_back
    Tp<class T>void sort(vector<T>&v){sort(all(v));}Tp<class T>void reverse(vector<T>&v){reverse(all(v));}
    Tp<class T>void unique(vector<T>&v){sort(all(v)),v.erase(unique(all(v)),v.end());}inline void reverse(string&s){reverse(s.begin(),s.end());}
    const int SZ=1<<23|233;
    struct FILEIN{char qwq[SZ],*S=qwq,*T=qwq,ch;
    #ifdef __WIN64
    #define GETC getchar
    #else
    inline char GETC(){return(S==T)&&(T=(S=qwq)+fread(qwq,1,SZ,stdin),S==T)?EOF:*S++;}
    #endif
    inline FILEIN&operator>>(char&c){while(isspace(c=GETC()));return*this;}inline FILEIN&operator>>(string&s){s.clear();while(isspace(ch=GETC()));if(!~ch)return*this;s=ch;while(!isspace(ch=GETC())&&~ch)s+=ch;return*this;}
    inline FILEIN&operator>>(char*str){char*cur=str;while(*cur)*cur++=0;cur=str;while(isspace(ch=GETC()));if(!~ch)return*this;*cur=ch;while(!isspace(ch=GETC())&&~ch)*++cur=ch;*++cur=0;return*this;}
    Tp<class T>inline void read(T&x){bool f=0;while((ch=GETC())<48&&~ch)f^=(ch==45);x=~ch?(ch^48):0;while((ch=GETC())>47)x=x*10+(ch^48);x=f?-x:x;}
    inline FILEIN&operator>>(int&x){return read(x),*this;}inline FILEIN&operator>>(ll&x){return read(x),*this;}inline FILEIN&operator>>(uint&x){return read(x),*this;}inline FILEIN&operator>>(ull&x){return read(x),*this;}
    inline FILEIN&operator>>(double&x){read(x);bool f=x<0;x=f?-x:x;if(ch^'.')return*this;double d=0.1;while((ch=GETC())>47)x+=d*(ch^48),d*=.1;return x=f?-x:x,*this;}
    }in;
    struct FILEOUT{const static int LIMIT=1<<22;char quq[SZ],ST[233];int sz,O,pw[233];
    FILEOUT(){set(7);rep(i,pw[0]=1,9)pw[i]=pw[i-1]*10;}~FILEOUT(){flush();}
    inline void flush(){fwrite(quq,1,O,stdout),fflush(stdout),O=0;}
    inline FILEOUT&operator<<(char c){return quq[O++]=c,*this;}inline FILEOUT&operator<<(string str){if(O>LIMIT)flush();for(char c:str)quq[O++]=c;return*this;}
    inline FILEOUT&operator<<(char*str){if(O>LIMIT)flush();char*cur=str;while(*cur)quq[O++]=(*cur++);return*this;}
    Tp<class T>void write(T x){if(O>LIMIT)flush();if(x<0){quq[O++]=45;x=-x;}do{ST[++sz]=x%10^48;x/=10;}while(x);while(sz)quq[O++]=ST[sz--];}
    inline FILEOUT&operator<<(int x){return write(x),*this;}inline FILEOUT&operator<<(ll x){return write(x),*this;}inline FILEOUT&operator<<(uint x){return write(x),*this;}inline FILEOUT&operator<<(ull x){return write(x),*this;}
    int len,lft,rig;void set(int l){len=l;}inline FILEOUT&operator<<(double x){bool f=x<0;x=f?-x:x,lft=x,rig=1.*(x-lft)*pw[len];return write(f?-lft:lft),quq[O++]='.',write(rig),*this;}
    }out;
    struct Math{
    vector<int>fac,inv;int mod;
    void set(int n,int Mod){fac.resize(n+1),inv.resize(n+1),mod=Mod;rep(i,fac[0]=1,n)fac[i]=fac[i-1]*i%mod;inv[n]=qpow(fac[n],mod-2);Rep(i,n-1,0)inv[i]=inv[i+1]*(i+1)%mod;}
    int qpow(int x,int y){int ans=1;for(;y;y>>=1,x=x*x%mod)if(y&1)ans=ans*x%mod;return ans;}int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*inv[m]%mod*inv[n-m]%mod;}
    int gcd(int x,int y){return!y?x:gcd(y,x%y);}int lcm(int x,int y){return x*y/gcd(x,y);}
    }math;
    // clang-format on
    
    int n, k;
    const int maxn = 2e5 + 52;
    set<int> s[maxn];
    
    struct suffix_auto_maton {
      int las, cnt;
    
      suffix_auto_maton() { las = cnt = 1; }
    
      int ch[maxn][26], fa[maxn], len[maxn];
      void ins(int c, int id) {
        int p = las, np = las = ++cnt;
        len[np] = len[p] + 1, s[np].insert(id);
        for (; p && !ch[p][c]; p = fa[p]) ch[p][c] = np;
        if (!p) {
          fa[np] = 1;
        } else {
          int q = ch[p][c];
          if (len[q] == len[p] + 1) {
            fa[np] = q;
          } else {
            int nq = ++cnt;
            memcpy(ch[nq], ch[q], sizeof(ch[q]));
            len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
            for (; p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq;
          }
        }
      }
    
      void ins(string s, int id) {
        las = 1;
        for (char c : s) {
          ins(c - 'a', id);
        }
      }
    } sam;
    
    int col[maxn];
    
    void merge(int u, int v) {
      if (s[u].size() < s[v].size()) swap(s[u], s[v]);
      for (int x : s[v]) s[u].insert(x);
      s[v].clear();
    }
    vector<int> g[maxn];
    void dfs(int u) {
      for (int v : g[u]) {
        dfs(v);
        merge(u, v);
      }
      col[u] = (int)s[u].size();
    }
    
    ll cnt[maxn];
    void dfs2(int u) {
      if (col[u] >= k) cnt[u] += sam.len[u] - sam.len[sam.fa[u]];
      for (int v : g[u]) {
        cnt[v] = cnt[u];
        dfs2(v);
      }
    }
    
    string str[maxn];
    signed main() {
      // code begin.
      in >> n >> k;
      rep(i, 1, n) {
        in >> str[i];
        sam.ins(str[i], i);
      }
      rep(i, 2, sam.cnt) g[sam.fa[i]].pb(i);
      dfs(1);
      dfs2(1);
      rep(i, 1, n) {
        int p = 1;
        ll ans = 0;
        for (char x : str[i]) {
          int c = x - 'a';
          p = sam.ch[p][c];
          ans += cnt[p];
        }
        out << ans << ' ';
      }
      return 0;
      // code end.
    }
    
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  • 原文地址:https://www.cnblogs.com/Isaunoya/p/12580714.html
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