Hackerrank University CodeSprint 2 Querying Sums on Strings

題目大意

令s,w 爲兩個字符串。
定義f(s,w,l,r) 表示w[l,r] 在s 中的出現次數。
現在給定串s ，m 對區間[li,ri] 和長度k ，需要回答q 個詢問，每個詢問給定一個長度爲k 的字符串w 和兩個a,b 要求：

∑i=abf(s,w,li,ri)

Data Constraint
n,m,k,q≤105,∑w≤105

題解

對k 進行分類討論。
當k>n√ 時，所有的詢問串最多隻有nn√ 個，所以直接暴力查詢，然後並查集維護一下即可。時間複雜度：O(nn√)
當k≤n√ 時，詢問串的長度不會很長，所以暴力枚舉每個詢問串的子串。時間複雜度：O(nk∗k2) 即O(nn√)

總的時間複雜度：O(nn√)

SRC

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std ;

#define N 300000 + 10
#define M 100000 + 10
typedef long long ll ;
struct Inter {
    int l , r ;
} In[M] ;
struct Query {
    int st ;
    int a , b ;
} P[M] ;
struct Note {
    int v , h ;
    Note ( int X = 0 , int Y = 0 ) { v = X , h = Y ; }
} tmp[N] ;

vector < Note > Que[M] ;

vector < int > Rec[325][325] ;

char str[M] , TS[M] ;
int S[N] ;
int Rank[N] , SA[N] , Height[N] ;
int tax[N] , tp[N] ;
int fa[N] , Num[N] ;
int maxc = 'z' ;
int n , m , Q , Qlen , Len ;
ll Ans[M] ;

bool cmp( int x , int y , int w ) { return tp[x] == tp[y] && tp[x+w] == tp[y+w] ; }

void Rsort() {
    memset( tax , 0 , sizeof(tax) ) ;
    for (int i = 1 ; i <= Len ; i ++ ) tax[Rank[tp[i]]] ++ ;
    for (int i = 1 ; i <= maxc ; i ++ ) tax[i] += tax[i-1] ;
    for (int i = Len ; i >= 1 ; i -- ) SA[tax[Rank[tp[i]]]--] = tp[i] ;
}

void Suffix() {
    for (int i = 1 ; i <= Len ; i ++ ) Rank[i] = S[i] , tp[i] = i ;
    Rsort() ;
    int p = 1 ;
    for (int w = 1 ; p < Len ; w += w , maxc = p ) {
        p = 0 ;
        for (int i = Len - w + 1 ; i <= Len ; i ++ ) tp[++p] = i ;
        for (int i = 1 ; i <= Len ; i ++ ) if ( SA[i] > w ) tp[++p] = SA[i] - w ;
        Rsort() ;
        swap( Rank , tp ) ;
        Rank[SA[1]] = p = 1 ;
        for (int i = 2 ; i <= Len ; i ++ ) {
            if ( !cmp( SA[i-1] , SA[i] , w ) ) p ++ ;
            Rank[SA[i]] = p ;
        }
    }
    int k = 0 ;
    for (int i = 1 ; i <= Len ; i ++ ) {
        if ( k ) k -- ;
        int j = SA[Rank[i]-1] ;
        while ( S[i+k] == S[j+k] ) k ++ ;
        Height[Rank[i]] = k ;
    }
}

bool cmp1( Note a , Note b ) { return a.v > b.v ; }

int Get( int x ) { return fa[x] == x ? x : fa[x] = Get(fa[x]) ; }

void Merge( int x , int y ) {
    int fx = Get(x) ;
    int fy = Get(y) ;
    if ( fx == fy ) return ;
    fa[fx] = fy ;
    Num[fy] += Num[fx] ;
}

void Solve0() {
    for (int i = 2 ; i <= Len ; i ++ ) {
        tmp[i-1].v = Height[i] ;
        tmp[i-1].h = i ;
    }
    sort( tmp + 1 , tmp + Len , cmp1 ) ;
    for (int i = 1 ; i <= Q ; i ++ ) {
        for (int j = P[i].a ; j <= P[i].b ; j ++ ) {
            int l = In[j].r - In[j].l + 1 ;
            Que[l].push_back( Note( i , j ) ) ;
        }
    }
    for (int i = 1 ; i <= Len ; i ++ ) {
        fa[i] = i ;
        if ( SA[i] <= n ) Num[i] = 1 ;
    }
    int k = 1 ;
    for (int i = Qlen ; i >= 1 ; i -- ) {
        while ( k < Len && tmp[k].v >= i ) {
            Merge( tmp[k].h - 1 , tmp[k].h ) ;
            k ++ ;
        }
        for (int j = 0 ; j < (signed)Que[i].size() ; j ++ ) {
            Note now = Que[i][j] ;
            int wz = P[now.v].st + In[now.h].l - 1 ;
            int fx = Get(Rank[wz]) ;
            Ans[now.v] += Num[fx] ;
        }
    }
    for (int i = 1 ; i <= Q ; i ++ ) printf( "%lld\n" , Ans[i] ) ;
}

bool flag[N] ;
int f[M][325] , g[325] ;
int Ord[N] , Sum[N] ;

void Solve1() {
    for (int i = 1 ; i <= m ; i ++ ) Rec[In[i].l][In[i].r].push_back(i) ;
    for (int i = 1 ; i <= Q ; i ++ ) {
        for (int j = 1 ; j <= Qlen ; j ++ ) flag[Rank[P[i].st+j-1]] = 1 ;
    }
    for (int i = 1 ; i <= Len ; i ++ ) Sum[i] = Sum[i-1] + (SA[i] <= n) ;
    for (int i = 1 ; i <= Len ; i ++ ) {
        int Maxv = 0 ;
        if ( flag[i] ) {
            Ord[i] = ++ Ord[0] ;
            for (int j = 0 ; j <= Qlen ; j ++ ) {
                f[Ord[i]][j] = Maxv + 1 ;
                Maxv = max( Maxv , g[j] ) ;
                g[j] = Maxv ;
            }
        } else {
            for (int j = 0 ; j <= Qlen ; j ++ ) {
                Maxv = max( Maxv , g[j] ) ;
                g[j] = Maxv ;
            }
        }
        if ( Height[i+1] <= Qlen ) g[Height[i+1]] = i ;
    }
    for (int i = 1 ; i <= Qlen ; i ++ ) g[i] = Len + 1 ;
    for (int i = Len ; i >= 1 ; i -- ) {
        int Minv = Len + 1 ;
        if ( flag[i] ) {
            for (int j = 0 ; j <= Qlen ; j ++ ) {
                f[Ord[i]][j] = Sum[Minv-1] - Sum[f[Ord[i]][j]-1] ;
                Minv = min( Minv , g[j] ) ;
                g[j] = Minv ;
            }
        } else {
            for (int j = 0 ; j <= Qlen ; j ++ ) {
                Minv = min( Minv , g[j] ) ;
                g[j] = Minv ;
            }
        }
        if ( Height[i] <= Qlen ) g[Height[i]] = i ;
    }
    for (int i = 1 ; i <= Q ; i ++ ) {
        ll ans = 0 ;
        for (int j = 1 ; j <= Qlen ; j ++ ) {
            for (int k = j ; k <= Qlen ; k ++ ) {
                int wz = P[i].st + j - 1 ;
                ll tot = f[Ord[Rank[wz]]][k-j+1] ;
                if ( !tot ) continue ;
                ll Cnt = upper_bound( Rec[j][k].begin() , Rec[j][k].end() , P[i].b ) - lower_bound( Rec[j][k].begin() , Rec[j][k].end() , P[i].a ) ;
                ans += tot * Cnt ;
            }
        }
        printf( "%lld\n" , ans ) ;
    }
}

int main() {
    scanf( "%d%d%d%d" , &n , &m , &Q , &Qlen ) ;
    Len = n ;
    scanf( "%s" , str + 1 ) ;
    for (int i = 1 ; i <= n ; i ++ ) S[i] = str[i] ;
    for (int i = 1 ; i <= m ; i ++ ) {
        scanf( "%d%d" , &In[i].l , &In[i].r ) ;
        In[i].l ++ , In[i].r ++ ;
    }
    int Qhead = 0 ;
    for (int i = 1 ; i <= Q ; i ++ ) {
        scanf( "%s" , TS + 1 ) ;
        scanf( "%d%d" , &P[i].a , &P[i].b ) ;
        P[i].a ++ , P[i].b ++ ;
        ++ Len , ++ maxc ;
        S[Len] = maxc ;
        P[i].st = Len + 1 ;
        for (int j = 1 ; j <= Qlen ; j ++ ) S[++Len] = TS[j] ;
    }
    Suffix() ;
    if ( Qlen > sqrt(n) ) Solve0() ;
    else Solve1() ;
    return 0 ;
}

以上.