【codeforces】741E. Arpa’s abnormal DNA and Mehrdad’s deep interest【后缀数组+分块】

题目链接：【codeforces】741E. Arpa’s abnormal DNA and Mehrdad’s deep interest

观察两个插入位置对他们rank的影响，可以发现，可以拆成求5段lcp，因此我们在将两个串拼接后求一个后缀数组，然后就可以直接对所有插入位置求出他们的rank了。

然后询问可以根据k分块，这题就做完了。
就是需要写一会儿。

#include <bits/stdc++.h>using namespace std ;typedef pair < int , int > pii ;const int MAXN = 200005 ;const int SQR = 300 ;struct Query {    int l , r , x , y , idx ;} ;/*===============sa================*/int sa[MAXN] , rnk[MAXN] , height[MAXN] ;int t1[MAXN] , t2[MAXN] , xy[MAXN] , c[MAXN] ;/*===============sa================*//*===============rmq================*/int dp[MAXN][18] ;pii f[MAXN][18] ;int logn[MAXN] ;/*===============rmq================*//*==============union===============*/pii val[MAXN] ;int p[MAXN] ;/*==============union===============*//*==============main===============*/char s1[MAXN] , s2[MAXN] ;int s[MAXN] , n , n1 , n2 ;int idx[MAXN] , ridx[MAXN] ;vector < Query > small[SQR] ;vector < pii > G[MAXN] ;pii ans[MAXN] ;/*==============main===============*/inline int comp ( int*r , int a , int b , int d ) {    return r[a] == r[b] && r[a + d] == r[b + d] ;}inline void get_height ( int n , int k = 0 ) {    for ( int i = 0 ; i <= n ; ++ i ) rnk[sa[i]] = i ;    for ( int i = 0 ; i < n ; ++ i ) {        if ( k ) -- k ;        int j = sa[rnk[i] - 1] ;        while ( s[i + k] == s[j + k] ) ++ k ;        height[rnk[i]] = k ;    }}inline int da ( int n , int m = 128 ) {    int *x = t1 , *y = t2 , i , d = 1 , p = 0 ;    for ( i = 0 ; i < m ; ++ i ) c[i] = 0 ;    for ( i = 0 ; i < n ; ++ i ) c[x[i] = s[i]] ++ ;    for ( i = 1 ; i < m ; ++ i ) c[i] += c[i - 1] ;    for ( i = n - 1 ; i >= 0 ; -- i ) sa[-- c[x[i]]] = i ;    for ( ; p < n ; d <<= 1 , m = p ) {        for ( p = 0 , i = n - d ; i < n ; ++ i ) y[p ++] = i ;        for ( i = 0 ; i < n ; ++ i ) if ( sa[i] >= d ) y[p ++] = sa[i] - d ;        for ( i = 0 ; i < m ; ++ i ) c[i] = 0 ;        for ( i = 0 ; i < n ; ++ i ) c[xy[i] = x[y[i]]] ++ ;        for ( i = 1 ; i < m ; ++ i ) c[i] += c[i - 1] ;        for ( i = n - 1 ; i >= 0 ; -- i ) sa[-- c[xy[i]]] = y[i] ;        swap ( x , y ) ;        p = 0 ;        x[sa[0]] = p ++ ;        for ( i = 1 ; i < n ; ++ i ) x[sa[i]] = comp ( y , sa[i - 1] , sa[i] , d ) ? p - 1 : p ++ ;    }    get_height ( n - 1 ) ;}inline void init_rmq ( int n ) {    for ( int i = 1 ; i <= n ; ++ i ) dp[i][0] = height[i] ;    logn[1] = 0 ;    for ( int i = 2 ; i <= n ; ++ i ) logn[i] = logn[i - 1] + ( i == ( i & -i ) ) ;    for ( int j = 1 ; ( 1 << j ) < n ; ++ j ) {        for ( int i = 1 ; i + ( 1 << j ) - 1 <= n ; ++ i ) {            dp[i][j] = min ( dp[i][j - 1] , dp[i + ( 1 << ( j - 1 ) )][j - 1] ) ;        }    }}inline int rmq ( int L , int R ) {    int k = logn[R - L + 1] ;    return min ( dp[L][k] , dp[R - ( 1 << k ) + 1][k] ) ;}inline int lcp ( int x , int y ) {    if ( x == y ) return n1 - x ;    x = rnk[x] , y = rnk[y] ;    if ( x > y ) swap ( x , y ) ;    return rmq ( x + 1 , y ) ;}inline int getid ( int x , int pos ) {    if ( pos < x ) return pos ;    if ( pos < x + n2 ) return n1 + 1 + pos - x ;    return pos - n2 ;}inline int cmp ( const int& x , const int& y ) {    int a[6] = { 0 , x , y , x + n2 , y + n2 , n1 + n2 } ;    sort ( a , a + 6 ) ;    for ( int i = 0 ; i < 5 ; ++ i ) {        int t1 = getid ( x , a[i] ) ;        int t2 = getid ( y , a[i] ) ;        int l = a[i] + lcp ( t1 , t2 ) ;        if ( l < a[i + 1] ) {            int t3 = getid ( x , l ) ;            int t4 = getid ( y , l ) ;            return s[t3] < s[t4] ;        }    }    return x < y ;}inline void build () {    n1 = strlen ( s1 ) ;    n2 = strlen ( s2 ) ;    n = n1 + n2 + 1 ;    for ( int i = 0 ; i < n1 ; ++ i ) {        s[i] = s1[i] ;    }    s[n1] = '#' ;    for ( int i = 0 ; i < n2 ; ++ i ) {        s[n1 + 1 + i] = s2[i] ;    }    s[n] = 0 ;    da ( n + 1 ) ;    init_rmq ( n ) ;    for ( int i = 0 ; i <= n1 ; ++ i ) {        idx[i] = i ;    }    sort ( idx , idx + n1 + 1 , cmp ) ;    for ( int i = 0 ; i <= n1 ; ++ i ) {        ridx[idx[i]] = i ;    }}inline void build_big_k () {    for ( int i = 0 ; i <= n1 ; ++ i ) f[i][0] = pii ( ridx[i] , i ) ;    logn[1] = 0 ;    for ( int i = 2 ; i <= n1 + 1 ; ++ i ) logn[i] = logn[i - 1] + ( i == ( i & -i ) ) ;    for ( int j = 1 ; ( 1 << j ) < n1 ; ++ j ) {        for ( int i = 0 ; i + ( 1 << j ) - 1 <= n1 ; ++ i ) {            f[i][j] = min ( f[i][j - 1] , f[i + ( 1 << ( j - 1 ) )][j - 1] ) ;        }    }}inline pii big_rmq ( int L , int R ) {    int k = logn[R - L + 1] ;    return min ( f[L][k] , f[R - ( 1 << k ) + 1][k] ) ;}inline void calc_big_k ( int k , Query x ) {    pii res ( MAXN , MAXN ) ;    for ( int i = 0 ; i <= n1 ; i += k ) {        if ( i + k <= x.l ) continue ;        if ( x.r < i ) break ;        int l = x.l < i ? x.x : max ( x.l % k , x.x ) ;        int r = i + k <= x.r ? x.y : min ( x.r % k , x.y ) ;        if ( l > r ) continue ;        res = min ( res , big_rmq ( i + l , i + r ) ) ;    }    ans[x.idx] = res ;}int F ( int x ) {    if ( p[x] == x ) return x ;    int lst = F ( p[x] ) ;    val[x] = min ( val[x] , val[p[x]] ) ;    return p[x] = lst ;}inline void calc_small_k ( int k , vector < Query >& op ) {    for ( int i = 0 ; i <= n1 ; ++ i ) {        p[i] = i ;        val[i] = pii ( ridx[i] , i ) ;        G[i].clear () ;    }    for ( int o = 0 ; o < op.size () ; ++ o ) {        Query x = op[o] ;        ans[x.idx] = pii ( MAXN , MAXN ) ;        for ( int i = x.x ; i <= x.y ; ++ i ) {            int l = x.l <= i ? 0 : ( x.l - i - 1 ) / k + 1 ;            int r = x.r < i ? -1 : ( x.r - i ) / k ;            if ( l > r ) continue ;            l = l * k + i ;            r = r * k + i ;            G[l].push_back ( pii ( r , x.idx ) ) ;        }    }    for ( int i = n1 ; i >= 0 ; -- i ) {        for ( int j = 0 ; j < G[i].size () ; ++ j ) {            int x = G[i][j].first , idx = G[i][j].second ;            F ( x ) ;            ans[idx] = min ( ans[idx] , val[x] ) ;        }        p[i] = i - k ;    }}inline void solve () {    build () ;    build_big_k () ;    Query x ;    int m , k ;    scanf ( "%d" , &m ) ;    for ( int i = 0 ; i < SQR ; ++ i ) small[i].clear () ;    for ( int i = 1 ; i <= m ; ++ i ) {        scanf ( "%d%d%d%d%d" , &x.l , &x.r , &k , &x.x , &x.y ) ;        x.idx = i ;        if ( k < SQR ) small[k].push_back ( x ) ;        else calc_big_k ( k , x ) ;    }    for ( int i = 1 ; i < SQR ; ++ i ) calc_small_k ( i , small[i] ) ;    for ( int i = 1 ; i <= m ; ++ i ) printf ( "%d%c" , ans[i].first == MAXN ? -1 : ans[i].second , i < m ? ' ' : '\n' ) ;}int main () {    while ( ~scanf ( "%s%s" , s1 , s2 ) ) solve () ;    return 0 ;}