题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2738
实在不想吐槽这个标题什么了，本来想找几道矩阵乘法的水题水水的，结果却成了裸数据结构。。。把X分成sqrt(n)个块，然后在没个块上面套棵主席树就可以啦~
空间复杂度：O( n ^ 2 log n )
时间复杂度：O( n ^ 2 log n + q sqrt( n ) log n )
代码：
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std ;
#define MAXN 510
#define L( t ) sgt[ t ].left
#define R( t ) sgt[ t ].right
#define S( t ) sgt[ t ].size
#define MAXV 10001000
#define MAXM 30
#define check( ch ) ( ch >= '0' && ch <= '9' )
void getint( int &t ) {
int ch ; for ( ch = getchar( ) ; ! check( ch ) ; ch = getchar( ) ) ;
t = ch - '0' ;
for ( ch = getchar( ) ; check( ch ) ; ch = getchar( ) ) {
t *= 10 , t += ch - '0' ;
}
}
int out[ 20 ] ;
void putint( int t ) {
if ( ! t ) putchar( '0' ) ; else {
out[ 0 ] = 0 ;
for ( ; t ; t /= 10 ) out[ ++ out[ 0 ] ] = t % 10 ;
while ( out[ 0 ] -- ) putchar( '0' + out[ out[ 0 ] + 1 ] ) ;
}
putchar( '\n' ) ;
}
struct node {
int left , right , size ;
node( ) {
left = right = size = 0 ;
}
} sgt[ MAXV ] ;
int V = 0 ;
void Add( int l , int r , int k , int u , int &t ) {
if ( ! t ) t = ++ V ;
S( t ) = S( u ) + 1 ;
if ( l == r ) return ;
int mid = ( l + r ) >> 1 ;
if ( k <= mid ) {
R( t ) = R( u ) ; Add( l , mid , k , L( u ) , L( t ) ) ;
} else {
L( t ) = L( u ) ; Add( mid + 1 , r , k , R( u ) , R( t ) ) ;
}
}
struct saver {
int x , y , v ;
bool operator < ( const saver &a ) const {
return v < a.v ;
}
} c[ MAXN * MAXN ] ;
int a[ MAXN ][ MAXN ] , n , N = 0 , b[ MAXN * MAXN ] , q , Rank[ MAXN ][ MAXN ] ;
int len ;
int Begin[ MAXM ] , End[ MAXM ] , M , block[ MAXM ][ MAXN ] , pre[ MAXN ][ MAXN ] ;
void Init( ) {
getint( n ) , getint( q ) ;
for ( int i = 0 ; i ++ < n ; ) {
for ( int j = 0 ; j ++ < n ; ) {
getint( a[ i ][ j ] ) ;
c[ ++ N ].v = a[ i ][ j ] ;
c[ N ].x = i , c[ N ].y = j ;
}
}
sort( c + 1 , c + N + 1 ) ;
N = 0 ;
for ( int i = 0 ; i ++ < n * n ; ) {
if ( i == 1 || c[ i ].v != c[ i - 1 ].v ) b[ ++ N ] = c[ i ].v ;
Rank[ c[ i ].x ][ c[ i ].y ] = N ;
}
len = int( sqrt( n ) ) ; M = n / len ;
for ( int i = 0 ; i ++ < M ; ) {
Begin[ i ] = ( i - 1 ) * len + 1 , End[ i ] = i * len ;
}
if ( M * len < n ) {
Begin[ M + 1 ] = M * len + 1 , End[ M + 1 ] = n ;
++ M ;
}
memset( pre , 0 , sizeof( pre ) ) ;
memset( block , 0 , sizeof( block ) ) ;
for ( int i = 0 ; i ++ < n ; ) {
for ( int j = 0 ; j ++ < n ; ) {
Add( 1 , N , Rank[ i ][ j ] , pre[ i ][ j - 1 ] , pre[ i ][ j ] ) ;
}
}
for ( int i = 0 ; i ++ < M ; ) {
for ( int j = 0 ; j ++ < n ; ) {
int temp = block[ i ][ j - 1 ] ;
for ( int k = Begin[ i ] ; k <= End[ i ] ; ++ k ) {
int rec = 0 ;
Add( 1 , N , Rank[ k ][ j ] , temp , rec ) ;
temp = rec ;
}
block[ i ][ j ] = temp ;
}
}
}
int range[ MAXN * MAXN ][ 2 ] , rn = 0 ;
void getrange( int x1 , int x2 , int y1 , int y2 ) {
int pos1 = ( x1 - 1 ) / len + 1 , pos2 = ( x2 - 1 ) / len + 1 ;
rn = 0 ;
if ( pos1 == pos2 ) {
if ( x1 == Begin[ pos1 ] && x2 == End[ pos1 ] ) {
range[ ++ rn ][ 0 ] = block[ pos1 ][ y2 ] ;
range[ rn ][ 1 ] = block[ pos1 ][ y1 - 1 ] ;
} else {
for ( int i = x1 ; i <= x2 ; ++ i ) {
range[ ++ rn ][ 0 ] = pre[ i ][ y2 ] ;
range[ rn ][ 1 ] = pre[ i ][ y1 - 1 ] ;
}
}
} else {
for ( int i = x1 ; i <= End[ pos1 ] ; ++ i ) {
range[ ++ rn ][ 0 ] = pre[ i ][ y2 ] ;
range[ rn ][ 1 ] = pre[ i ][ y1 - 1 ] ;
}
for ( int i = Begin[ pos2 ] ; i <= x2 ; ++ i ) {
range[ ++ rn ][ 0 ] = pre[ i ][ y2 ] ;
range[ rn ][ 1 ] = pre[ i ][ y1 - 1 ] ;
}
if ( ( ++ pos1 ) <= ( -- pos2 ) ) {
for ( int i = pos1 ; i <= pos2 ; ++ i ) {
range[ ++ rn ][ 0 ] = block[ i ][ y2 ] ;
range[ rn ][ 1 ] = block[ i ][ y1 - 1 ] ;
}
}
}
}
int query( int x1 , int x2 , int y1 , int y2 , int k ) {
getrange( x1 , x2 , y1 , y2 ) ;
int l = 1 , r = N ;
while ( l < r ) {
int mid = ( l + r ) >> 1 , sum = 0 ;
for ( int i = 0 ; i ++ < rn ; ) {
sum += S( L( range[ i ][ 0 ] ) ) , sum -= S( L( range[ i ][ 1 ] ) ) ;
}
if ( k <= sum ) {
r = mid ;
for ( int i = 0 ; i ++ < rn ; ) {
range[ i ][ 0 ] = L( range[ i ][ 0 ] ) ;
range[ i ][ 1 ] = L( range[ i ][ 1 ] ) ;
}
} else {
l = mid + 1 , k -= sum ;
for ( int i = 0 ; i ++ < rn ; ) {
range[ i ][ 0 ] = R( range[ i ][ 0 ] ) ;
range[ i ][ 1 ] = R( range[ i ][ 1 ] ) ;
}
}
}
return b[ l ] ;
}
void Solve( ) {
while ( q -- ) {
int x1 , y1 , x2 , y2 , k ;
getint( x1 ) , getint( y1 ) , getint( x2 ) , getint( y2 ) ;
getint( k ) ;
putint( query( x1 , x2 , y1 , y2 , k ) ) ;
}
}
int main( ) {
Init( ) ;
Solve( ) ;
return 0 ;
}
