POJ 2823 castrated line segment tree

The better method for this question is monotonous queue. Implement it later. Use castrated line tree to write 9600 ms + + records ~

The simple line segment tree will not be mentioned. For details, you can read previous articles or the following code .....

Although it's a bit rough to count ....

#include<iostream>#define MAXN 1000001using namespace std;int _max[MAXN<<2];int _min[MAXN<<2];int mx( int a,int b ){    return a>b?a:b;}int mn( int a,int b ){    return a<b?a:b;}void PushUp( int rt ){     _max[rt]=mx( _max[rt<<1],_max[rt<<1|1] );     _min[rt]=mn( _min[rt<<1],_min[rt<<1|1] );}void build(int l,int r,int rt){     if( l==r ){         scanf( "%d",&_max[rt] );         _min[rt]=_max[rt];         return ;     }     int m=(l+r)>>1;     build( l,m,rt<<1 );     build( m+1,r,rt<<1|1 );     PushUp(rt);}int query( int l,int r,int c,int L,int R,int rt ){    if( l<=L&&R<=r )    {        if( c )        return _max[rt];        else        return _min[rt];    }    int m=(L+R)>>1;    int res;    if( c )    {         res=-214748363;        if( l<=m ) res=mx(res,query( l,r,c,L,m,rt<<1) );        if( m<r ) res=mx( res,query( l,r,c,m+1,R,rt<<1|1) );        return res;    }    else    {        res=214748363;        //res=100;        if( l<=m ) res=mn(res,query( l,r,c,L,m,rt<<1) );        if( m<r ) res=mn( res,query( l,r,c,m+1,R,rt<<1|1) );        return res;    }}int main(){    int n,c,i;    while( scanf( "%d %d",&n,&c)!=EOF )    {           build(1,n,1);           printf( "%d",query( 1,c,0,1,n,1 ) );           for( i=2;;i++ )           {                if( i+c-1<=n ){                    printf( " %d",query( i,i+c-1,0,1,n,1 ) );                }                else                    break;           }           printf( "\n%d",query( 1,c,1,1,n,1 ) );           for( i=2;;i++ )           {                if( i+c-1<=n ){                    printf( " %d",query( i,i+c-1,1,1,n,1 ) );                }                else                    break;           }           printf( "\n" );    }    return 0;}