AC's code , FZU 2011年3月月賽之 C, FZU 2012
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
AC is given two strings A and B, and then he wants to insert B into A. As we know, there are |A| + 1 possible ways to make the final string! For example, if A = “orz” and B = “p”, then AC may get four final strings: “porz”, “oprz”, “orpz”, “orzp”. However, these strings are not so beautiful, so AC sort them in lexicographic order, then he gets {“oprz”, “orpz”, “orzp”,”porz”}.
Now AC wants to know the k-th string in the sorted strings! For example, if AC wants to know the first string in the sorted strings, the answer is “oprz”, the second is “orpz”, the third is “orzp”, and the 4-th is “porz”.
Now you are given string A and B, and the number K, in this problem, AC guarantees that K is valid.
Input
The first line of the input data is an integer number T (T <= 1000), represent the number of test cases.
For each case, three lines follow.
The first line has one string indicates A. (Without any space in A)
The second line has one string indicates B. (Without any space in B)
The third line has one integer indicates K.
(1 <= |A|, |B| <= 10000, 1 <= K <= |A| + 1)
Output
For each test case, output a single line “Case idx:” where idx is the test case number starts from 1. Then one line output the k-th string in the sorted string.
Sample Input
orz
p
1
orz
p
2
orz
p
3
orz
p
4
Sample Output
oprz
Case 2:
orpz
Case 3:
orzp
Case 4:
porz
分析:
字符串 hash,求第 k 小元素。
字符串hash 函數為
hash ( s[ 0..n ] ) = ( s[0]*q^n + s[1]*q^(n-1) + ...... + s[n-1]*q^1 + s[n]*q^0 ) mod p,其中,p, q 為大質數,
適當預處理(見函數init )后,B 插入在 A 中任意位置所形成的字符串的任意前綴的hash 值都能 O(1) 計算出來(見函數hash)。
求第 k 小元素可以使用修改的快速排序(見函數sort),其中要注意使用隨機,否則超時。復雜度 O(n)。
比較兩個字符串大小時,只比較它們的最長公共前綴的后一個字符(見函數 cmp)。
而最長公共前綴通過二分其長度,用hash值判斷是否相等,可以O(log(|A|+|B|) )得到(見函數 cmp)。
另一種做法是用后綴數組的。
我的代碼:
#include <stdio.h>2
#include <string.h>3
#include <stdlib.h>4
#include <time.h>5

6
typedef long long Lint;7

8
#define PRIME 555666779
#define HH 2345678910

11
#define L 2100912

13
int la, lb, k;14
char a[ L ], b[ L ], c[ L + L ];15
Lint ha[ L ], hb[ L ], hh[ L ];16

17

void init()
{18
int i;19
srand( time( 0 ) );20

21
memset( ha, 0, sizeof(ha) );22
ha[ 0 ] = a[ 0 ] % PRIME;23

for ( i = 1; a[ i ]; ++i )
{24
ha[ i ] = ( ha[ i - 1 ] * HH + a[ i ] ) % PRIME;25
}26
la = i;27

28
memset( hb, 0, sizeof(hb) );29
hb[ 0 ] = b[ 0 ] % PRIME;30

for ( i = 1; b[ i ]; ++i )
{31
hb[ i ] = ( hb[ i - 1 ] * HH + b[ i ] ) % PRIME;32
}33
lb = i;34

35
hh[ 0 ] = 1;36

for ( i = 1; i < L; ++i )
{37
hh[ i ] = ( hh[ i -1 ] * HH ) % PRIME;38
}39
}40

41

/**//*42
insert b after a[ i ]43
query hash of [0..j]44
*/45

Lint hash( int i, int j )
{46

if ( i < 0 )
{47

if ( j < lb )
{48
return hb[ j ];49
}50
return ( hb[ lb-1 ] * hh[ j-lb+1 ] + ha[ j-lb ] ) % PRIME;51
}52

if ( j <= i )
{53
return ha[ j ];54
}55

if ( j <= i + lb )
{56
return ( ha[ i ] * hh[ j-i ] + hb[ j-i-1 ] ) % PRIME;57
}58
return ( ha[i]*hh[j-i] + hb[lb-1]*hh[j-i-lb] + (ha[j-lb]+PRIME-((ha[i]*hh[j-lb-i])%PRIME)) ) % PRIME;59
}60

61

char get( int i, int j )
{62

if ( i < 0 )
{63

if ( j < lb )
{64
return b[ j ];65
}66
return a[ j-lb ];67
}68

if ( j <= i )
{69
return a[ j ];70
}71

if ( j <= i + lb )
{72
return b[ j-i-1 ];73
}74
return a[ j-lb ];75
}76

77

int cmp( int i, int j )
{78
int low = 0, high = la+lb-1, mid, lcp=-1;79

while ( low <= high )
{80
mid = ( low + high ) / 2;81

if ( hash(i,mid)==hash(j,mid) )
{82
low = mid + 1;83

if ( lcp < mid )
{84
lcp = mid;85
}86
}87

else
{88
high = mid - 1;89
}90
}91
return get(i,lcp+1) - get(j,lcp+1);92
}93

94
int idx[ L ];95

void sort( int le, int ri, int k )
{96
int i, j, x;97

if ( (le>=ri) || (k<1) )
{98
return;99
}100
i = le + ( rand() % ( ri - le + 1 ) );101
x = idx[ i ];102
idx[ i ] = idx[ le ];103
idx[ le ] = x;104
i = le;105
j = ri;106

while ( i < j )
{107

while ( (i<j) && (cmp(x,idx[j])<=0) )
{108
--j;109
}110

if ( i < j )
{111
idx[ i++ ] = idx[ j ];112
}113

while ( (i<j) && (cmp(idx[i],x)<=0) )
{114
++i;115
}116

if ( i < j )
{117
idx[ j-- ] = idx[ i ];118
}119
}120
idx[ i ] = x;121

if ( i - le + 1 < k )
{122
sort( i+1, ri, k-(i-le+1) );123
}124

else if ( i - le + 1 > k )
{125
sort( le, i-1, k );126
}127
}128

129

char* solve()
{130
int i, p;131
char tmp;132

for ( i = 0; i <= la; ++i )
{133
idx[ i ] = i - 1;134
}135
sort( 0, la, k );136
c[ 0 ] = 0;137
p = idx[ k - 1 ];138
tmp = a[ p + 1 ];139
a[ p + 1 ] = 0;140
strcat( c, a );141
a[ p + 1 ] = tmp;142
strcat( c, b );143
strcat( c, a+p+1 );144
return (char*)c;145
}146

147

int main()
{148
int td, cd = 0;149
scanf( "%d", &td );150

while ( td-- > 0 )
{151
scanf( "%s%s%d", a, b, &k );152
init();153
printf( "Case %d:\n%s\n", ++cd, solve() );154
}155
return 0;156
}157

posted on 2011-03-21 21:12 coreBugZJ 閱讀(1902) 評論(9) 編輯 收藏 引用 所屬分類: ACM

