hdu 3553 Just a String (后缀数组)
hdu 3553 Just a String (后缀数组)
题意：很简单，问一个字符串的第k大的子串是谁。
解题思路：后缀数组。先预处理一遍，把能算的都算出来。将后缀按sa排序，假如我们知道答案在那个区间范围内了（假设为[l,r]），那么我们算下这个区间内的lcp的最小值（设最小值的位置为mid，大小为x），如果x*(r-l+1)>=k，那么，答案就是这个区间的lcp的最小值的某一部分（具体是哪一部分，画个图稍微算下就出来了）。如果x * ( r - l + 1 ) < k 那么我们分两种情况考虑，如果[l,mid]区间范围内的字符串总数大于等于k，那么把区间范围缩小到[l,mid]，否则范围缩小到[mid+1,r]。一点点的逼近答案就可以了。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll __int64
using namespace std ;
const int maxn = 111111 ;
int min ( int a , int b ) { return a < b ? a : b ; }
int f[maxn] ;
int dp[22][maxn] ;
ll sum[maxn] ;
char s1[maxn] ;
int s[maxn] ;
struct Suf{
int wa[maxn] , wb[maxn] , ws[maxn] , wv[maxn] ;
int rank[maxn] , hei[maxn] , sa[maxn] ;
int cmp ( int *r , int i , int j , int l ){ return r[i] == r[j] && r[i+l] == r[j+l] ; }
void da ( int *r , int n , int m ){
int *x = wa , *y = wb , *t ;
int i , j , k , p ;
for ( i = 0 ; i < m ; i ++ ) ws[i] = 0 ;
for ( i = 0 ; i < n ; i ++ ) ws[x[i]=r[i]] ++ ;
for ( i = 1 ; i < m ; i ++ ) ws[i] += ws[i-1] ;
for ( i = n - 1 ; i >= 0 ; i -- ) sa[--ws[x[i]]] = i ;
for ( j = 1 , p = 1 ; p < n ; j *= 2 , m = p ) {
for ( p = 0 , i = n - j ; i < n ; i ++ ) y[p++] = i ;
for ( i = 0 ; i < n ; i ++ ) if ( sa[i] >= j ) y[p++] = sa[i] - j ;
for ( i = 0 ; i < m ; i ++ ) ws[i] = 0 ;
for ( i = 0 ; i < n ; i ++ ) ws[x[i]] ++ ;
for ( i = 1 ; i < m ; i ++ ) ws[i] += ws[i-1] ;
for ( i = n - 1 ; i >= 0 ; i -- ) sa[--ws[x[y[i]]]] = y[i] ;
for ( t = x , x = y , y = t ,x[sa[0]] = 0 , p = 1 , i = 1 ; i < n ; i ++ )
x[sa[i]] = cmp ( y , sa[i-1] , sa[i] , j ) ? p - 1 : p ++ ;
}
k = 0 ;
for ( i = 1 ; i < n ; i ++ ) rank[sa[i]] = i ;
for ( i = 0 ; i < n - 1 ; hei[rank[i++]] = k )
for ( k ? k -- : 0 , j = sa[rank[i]-1] ; r[i+k] == r[j+k] ; k ++ ) ;
}
int min_hei ( int x , int y ) {
return ( hei[x] < hei[y] ? x : y ) ;
}
void rmq ( int n ) {
int i , j ;
for ( i = 1 ; i <= n ; i ++ ) dp[0][i] = i ;
for ( i = 1 ; i <= 20 ; i ++ )
for ( j = 1 ; j + ( 1 << i ) - 1 <= n ; j ++ )
dp[i][j] = min_hei ( dp[i-1][j] , dp[i-1][j+(1<<(i-1))] ) ;
}
int query ( int l , int r ) {
if ( l > r ) swap ( l , r ) ;
l ++ ;//要从height[l+1]到height[r]之间求最小值
if ( l == r ) return dp[0][l] ;
int k = r - l + 1 ;
return min_hei ( dp[f[k]][l] , dp[f[k]][r-(1<<f[k])+1] ) ;
}
void solve ( int n , ll k ) {
rmq ( n ) ;
int l = 1 , r = n , i;
sum[0] = 0 ;
for ( i = 1 ; i <= n ; i ++ )
sum[i] = sum[i-1] + n - sa[i] ;
int h = 0 ;
int pos = 0 , len ;
while ( l < r ) {
int mid = query ( l , r ) - 1 ;
// printf ( "l = %d , r = %d mid = %d , k = %I64d , fuck = %d\n" , l , r , mid , k , ( hei[mid] - h ) * ( r - l + 1 ) ) ;
if ( k <= (ll) ( hei[mid+1] - h ) * ( r - l + 1 ) ) {
pos = l ;
len = h + k / ( r - l + 1 ) + ( k % ( r - l + 1 ) != 0 ) ;
// printf ( "pos = %d , l = %d\n" , pos , len ) ;
break ;
}
k -= (ll) (hei[mid+1] - h ) * ( r - l + 1 ) ;
if ( k <= sum[mid] - sum[l-1] - (ll) hei[mid+1] * ( mid - l + 1 ) ) {
r = mid ;
}
else {
k -= sum[mid] - sum[l-1] - (ll) hei[mid+1] * ( mid - l + 1 ) ;
l = mid + 1 ;
}
h = hei[mid+1] ;
}
if ( !pos ) pos = l , len = h + k ;
for ( i = 0 ; i < len ; i ++ )
printf ( "%c" , s[sa[pos]+i] ) ;
puts ( "" ) ;
}
} arr ;
int main () {
int cas , i , j , ca = 0 ;
ll m ;
j = 0 ;
for ( i = 1 ; i < maxn - 1111 ; i ++ ) {
if ( i > 1 << j + 1 ) j ++ ;
f[i] = j ;
}
scanf ( "%d" , &cas ) ;
while ( cas -- ) {
scanf ( "%s" , s1 ) ;
scanf ( "%I64d" , &m ) ;
int len = strlen ( s1 ) ;
for ( i = 0 ; i < len ; i++ ) s[i] = s1[i] ;
s[len] = 0 ;
arr.da ( s , len + 1 , 411 ) ;
printf ( "Case %d: " , ++ ca ) ;
arr.solve ( len , m ) ;
}
}
/*
10000
ddff 9
*/
一些数据：
10000
dDFdwb78648 50
DDFddwFd77866886 50
ddffddff66555566 50
ddff66555566 50
66555566 25
ddffddff 25
ddff 9
ABC 2
BBC 3