poj 3264 -- Balanced Lineup (区间最值，线段树/RMQ）

最新推荐文章于 2023-10-07 08:26:33 发布

原创最新推荐文章于 2023-10-07 08:26:33 发布 · 734 阅读

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本文介绍了如何使用预处理最小最大值表和线段树两种方法解决区间最值查询（RMQ）问题。通过具体代码实现展示了每种方法的构建过程及查询操作，为读者提供了理解RMQ问题解决方案的途径。

第一次RMQ ，还好

RMQ版：

# include <cstdio>
# include <iostream>
# include <set>
# include <map>
# include <vector>
# include <list>
# include <queue>
# include <stack>
# include <cstring>
# include <string>
# include <cstdlib>
# include <cmath>
# include <algorithm>

using namespace std ;

const int maxn = 51000 ;

int dp1 [ 20 ] [ maxn ] ;
int dp2 [ 20 ] [ maxn ] ;                //小的放前面效率高不少
int a [ 51000 ] ;
int n ;

void getdp ( )
{
    for ( int i = 1 ; i <= n ; i ++ )
    {
        dp1 [ 0 ] [ i ] = dp2 [ 0 ] [ i ] = a [ i ] ;
    }
    int len = floor ( log2 ( 51000 ) ) ;       //C++没有log2，取自然对数算
    for ( int j = 1 ; j <= len ; j ++ )
        for ( int i = 1 ; i + ( 1 << j ) - 1 <= n ; i ++ )
        {
            dp1 [ j ] [ i ] = min ( dp1 [ j - 1 ] [ i ] , dp1 [ j - 1 ] [ i + (1 << (j - 1)) ] ) ;
            dp2 [ j ] [ i ] = max ( dp2 [ j - 1 ] [ i ] , dp2 [ j - 1 ] [ i + (1 << (j - 1)) ] ) ;
        }
}

int main ( )
{
    int q ;
    scanf ( "%d%d" , & n , & q ) ;
    for ( int i = 1 ; i <= n ; i ++ )
    {
        scanf ( "%d" , a + i ) ;
    }
    getdp ( ) ;
    while ( q -- )
    {
        int m , n ;
        scanf ( "%d%d" , & m , & n ) ;
        int len = floor ( log2 ( n - m + 1 ) ) ;
        int maxx = max ( dp2 [ len ] [ m ] , dp2 [ len ] [ n - ( 1 << len ) + 1 ] ) ;
        int minn = min ( dp1 [ len ] [ m ] , dp1 [ len ] [ n - ( 1 << len ) + 1 ] ) ;
        printf ( "%d\n" , maxx - minn ) ;
    }
}

线段树版：

# include <cstdio>
# include <iostream>
# include <set>
# include <map>
# include <vector>
# include <list>
# include <queue>
# include <stack>
# include <cstring>
# include <string>
# include <cstdlib>
# include <cmath>
# include <algorithm>

using namespace std ;

const int maxn = 51000 ;
const int MAX = 2100000000 ;

int low , high ;

struct Tree
{
    int maxx , minn ;
} tree [ maxn * 4 ] ;

void pushup ( int pos )
{
    tree [ pos ] . minn = min ( tree [ pos * 2 ] . minn , tree [ pos * 2 + 1 ] . minn ) ;
    tree [ pos ] . maxx = max ( tree [ pos * 2 ] . maxx , tree [ pos * 2 + 1 ] . maxx ) ;
}

void build ( int l , int r , int pos )
{
    if ( l == r )
    {
        scanf ( "%d" , & tree [ pos ] . minn ) ;
        tree [ pos ] . maxx = tree [ pos ] . minn ;
        return ;
    }
    int m = ( l + r ) / 2 ;
    build ( l , m , pos * 2 ) ;
    build ( m + 1 , r , pos * 2 + 1 ) ;
    pushup ( pos ) ;
}

void quer ( int l , int r , int pos , int L , int R )
{
    if ( l >= L && r <= R )
    {
        high = max ( high , tree [ pos ] . maxx ) ;
        low = min ( low , tree [ pos ] . minn ) ;
        return ;
    }
    int m = ( l + r ) / 2 ;
    if ( m >= L )
        quer ( l , m , pos * 2 , L , R ) ;
    if ( m + 1 <= R )
        quer ( m + 1 , r , pos * 2 + 1 , L , R ) ;
}

int main ( )
{
    int n , q ;
    scanf ( "%d%d" , & n , & q ) ;
    build ( 1 , n , 1 ) ;
    while ( q -- )
    {
        int L , R ;
        scanf ( "%d%d" , & L , & R ) ;
        high = 0 , low = MAX ;
        quer ( 1 , n , 1 , L , R ) ;
        printf ( "%d\n" , high - low ) ;
    }
}