【codeforces】741E. Arpa’s abnormal DNA and Mehrdad’s deep interest【后缀数组+分块】
2016-12-09 14:36
519 查看
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。
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 = 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 ; }
相关文章推荐
- Codeforces 452E Three strings 后缀数组 + 并查集
- CodeForces - 432D 后缀数组
- 【BZOJ2119】股市的预测 后缀数组+分块
- CodeForces 128B String 后缀数组 或 优先队列维护 求第K小子串
- POJ - 3693 Maximum repetition substring 后缀数组 分块
- Codeforces Good Bye 2015 D. New Year and Ancient Prophecy 后缀数组 树状数组 dp
- codeforces 427D Match & Catch(后缀数组,字符串)
- _bzoj1031 [JSOI2007]字符加密Cipher【后缀数组】
- POJ2406:Power Strings(后缀数组DC3)
- 【UOJ #35】后缀排序 后缀数组模板
- !CodeForces 543A Writing Code --DP--(三维dp,滚动数组)
- B-树、后缀数组、网络流算法、问题规约、不可解性
- CF(427D-Match & Catch)后缀数组应用
- [ACM] hdu 5147 Sequence II (树状数组,前缀和,后缀和)
- Hdu-5769 Substring (SA后缀数组)
- Hdu1556 Color the ball [分块][树状数组]
- 后缀数组:最长重复子串 (marked)
- 后缀数组及其应用
- Codeforces 570E,DP+滚动数组
- poj 2217 后缀数组求最长公共子串