• P2414 [NOI2011]阿狸的打字机 [AC自动机,fail树,dfs序,树状数组]


    板子,不说了/fad

    // 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;
    #define Tp template
    using pii = pair<int, int>;
    #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()); }
    const int SZ = 1 << 23 | 233;
    struct FILEIN { char qwq[SZ], *S = qwq, *T = qwq, ch;
    #ifdef __WIN64
    #define GETC getchar
    #else
      char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
    #endif
      FILEIN& operator>>(char& c) {while (isspace(c = GETC()));return *this;}
      FILEIN& operator>>(string& s) {while (isspace(ch = GETC())); s = ch;while (!isspace(ch = GETC())) s += ch;return *this;}
      Tp<class T> void read(T& x) { bool sign = 0;while ((ch = GETC()) < 48) sign ^= (ch == 45); x = (ch ^ 48);
        while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48); x = sign ? -x : x;
      }FILEIN& operator>>(int& x) { return read(x), *this; } FILEIN& operator>>(ll& x) { return read(x), *this; }
    } in;
    struct FILEOUT {const static int LIMIT = 1 << 22 ;char quq[SZ], ST[233];int sz, O;
      ~FILEOUT() { flush() ; }void flush() {fwrite(quq, 1, O, stdout); fflush(stdout);O = 0;}
      FILEOUT& operator<<(char c) {return quq[O++] = c, *this;}
      FILEOUT& operator<<(string str) {if (O > LIMIT) flush();for (char c : str) quq[O++] = c;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--];
      }FILEOUT& operator<<(int x) { return write(x), *this; } FILEOUT& operator<<(ll x) { return write(x), *this; }
    } out;
    #define int long long
    
    int q ;
    const int maxn = 1e5 + 51 ;
    string s ;
    int ed[maxn] , id = 0 ;
    int cpy[maxn][26] ;
    vector < int > g[maxn] ;
    struct ACAM {
    	int ch[maxn][26] , cnt = 1 , fa[maxn] , fail[maxn] ;
    	void build() {
    		int p = 1 ;
    		for(char c : s) 
    			if('a' <= c && c <= 'z') {
    				if(! ch[p][c - 'a']) fa[ch[p][c - 'a'] = ++ cnt] = p ;
    				p = ch[p][c - 'a'] ;
    			}
    			else if(c == 'P') ed[++ id] = p ;
    			else if(c == 'B') p = fa[p] ;
    		memcpy(cpy , ch , sizeof(ch)) ;
    		queue <int> q ;
    		for(int i = 0 ; i < 26 ; i ++)
    			if(ch[1][i])
    				fail[ch[1][i]] = 1 , q.push(ch[1][i]) ;
    			else 
    				ch[1][i] = 1 ;
    		while(! q.empty()) {
    			int u = q.front() ;
    			q.pop() ;
    			for(int i = 0 ; i < 26 ; i ++)
    				if(ch[u][i]) 
    					fail[ch[u][i]] = ch[fail[u]][i] , q.push(ch[u][i]) ;
    				else 
    					ch[u][i] = ch[fail[u]][i] ;
    		}
    		for(int i = 2 ; i <= cnt ; i ++)
    			g[fail[i]].pb(i) ;
    	}
    } acam ;
    
    int L[maxn] , R[maxn] , idx = 0 ;
    void dfs(int u) {
    	L[u] = ++ idx ;
    	for(int v : g[u])
    		dfs(v) ;
    	R[u] = idx ;
    }
    
    int ans[maxn] ;
    vector < pii > qr[maxn] ;
    
    struct BIT {
    	int c[maxn + 5] ;
    	int low(int x) {
    		return x & -x ;
    	}
    	void add(int x , int y) {
    		for( ; x <= maxn ; x += low(x))
    			c[x] += y ;
    	}
    	int qry(int x) {
    		int ans = 0 ;
    		for( ; x ; x ^= low(x))
    			ans += c[x] ;
    		return ans ;
    	}
    } bit ;
    
    void solve(int u) {
    	bit.add(L[u] , 1) ;
    	for(auto x : qr[u])
    		ans[x.second] = bit.qry(R[ed[x.first]]) - bit.qry(L[ed[x.first]] - 1) ;
    	for(int i = 0 ; i < 26 ; i ++)
    		if(cpy[u][i])
    			solve(cpy[u][i]) ;
    	bit.add(L[u] , -1) ;
    }
    
    signed main() {
      // code begin.
    	in >> s ;
    	in >> q ;
    	acam.build() ;
    	dfs(1) ;
    	for(int i = 1 ; i <= q ; i ++) {
    		int a , b ;
    		in >> a >> b ;
    		qr[ed[b]].pb(a , i) ;
    	}
    	solve(1) ;
    	for(int i = 1 ; i <= q ; i ++)
    		out << ans[i] << '
    ' ;
    	return 0;
      // code end.
    }
    
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  • 原文地址:https://www.cnblogs.com/Isaunoya/p/12384231.html
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