HDU6579 Operation

题目链接

问题分析

区间求异或和最大,比较自然的想到了线性基。而每次求一个区间的线性基显然是行不通的。我们考虑在每个位置求出首位置到当前位置的线性基。同时我们要使线性基中高位的位置所选的数尽量靠后。这样我们维护线性基的时候在同时维护一个位置信息就好了。

参考程序

#include <bits/stdc++.h>
//#define Debug
using namespace std;

const int Maxn = 1000010;
const int MaxBit = 30;
int n, m, A[ Maxn ], BitBase[ Maxn ][ MaxBit ], Pos[ Maxn ][ MaxBit ];

void Work();
int main() {
	int TestCases;
	scanf( "%d", &TestCases );
	for( ; TestCases--; ) Work();
	return 0;
}

void Work() {
	scanf( "%d%d", &n, &m );
	for( int i = 1; i <= n; ++i ) scanf( "%d", A + i );
	memset( BitBase, 0, sizeof( BitBase ) );
	memset( Pos, 0, sizeof( Pos ) );
	for( int i = 1; i <= n; ++i ) {
		for( int j = 0; j < MaxBit; ++j ) BitBase[ i ][ j ] = BitBase[ i - 1 ][ j ], Pos[ i ][ j ] = Pos[ i - 1 ][ j ];
		int Position = i;
		for( int j = MaxBit - 1; j >= 0; --j ) 
			if( ( A[ i ] >> j ) & 1 ) 
				if( BitBase[ i ][ j ] ) {
					if( Pos[ i ][ j ] < Position ) {
						swap( BitBase[ i ][ j ], A[ i ] );
						swap( Pos[ i ][ j ], Position );
					}
					A[ i ] ^= BitBase[ i ][ j ];
				} else {
					BitBase[ i ][ j ] = A[ i ];
					Pos[ i ][ j ] = Position;
					break;
				}
	}
	#ifdef Debug
		for( int i = 1; i <= n; ++i ) {
			for( int j = 0; j < MaxBit; ++j ) printf( "(%d,%d), ", BitBase[ i ][ j ], Pos[ i ][ j ] );
			printf( "\n" );
		}
	#endif
	int LastAns = 0;
	for( int i = 1; i <= m; ++i ) {
		int Opt; scanf( "%d", &Opt );
		if( Opt == 1 ) {
			++n; scanf( "%d", A + n );
			A[ n ] ^= LastAns;
			for( int j = 0; j < MaxBit; ++j ) BitBase[ n ][ j ] = BitBase[ n - 1 ][ j ], Pos[ n ][ j ] = Pos[ n - 1 ][ j ];
			int Position = n;
			for( int j = MaxBit - 1; j >= 0; --j )
				if( ( A[ n ] >> j ) & 1 )
					if( !BitBase[ n ][ j ] ) {
						BitBase[ n ][ j ] = A[ n ];
						Pos[ n ][ j ] = Position;
						break;
					} else {
						if( Pos[ n ][ j ] < Position ) {
							swap( BitBase[ n ][ j ], A[ n ] );
							swap( Pos[ n ][ j ], Position );
						}
						A[ n ] ^= BitBase[ n ][ j ];
					}
		} else {
			int l, r; scanf( "%d%d", &l, &r );
			l = ( l ^ LastAns ) % n + 1;
			r = ( r ^ LastAns ) % n + 1;
			if( l > r ) swap( l, r );
			LastAns = 0;
			for( int j = MaxBit - 1; j >= 0; --j )
				if( ( ( LastAns >> j ) & 1 ) == 0 && Pos[ r ][ j ] >= l )
					LastAns ^= BitBase[ r ][ j ];
			printf( "%d\n", LastAns );
		}
	}
	return;
}
posted @ 2019-09-08 15:19  chy_2003  阅读(...)  评论(...编辑  收藏