BZOJ 1251 FHQ_Treap平衡树

题目链接

题意:

给一个长度为n初始全为0的数组,维护三种操作:

1 l r x:给区间[l,r]上每个元素都加上x

2 l r:反转区间[l,r](比如:[1,2,3]反转后变成[3,2,1])

3 l r:输出区间[l,r]的最大值

思路:

FHQ_Treap平衡树的多种操作

C++代码:

#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<string>
#include<cstdio>
#include<cstring>
#include<complex>
#include<iostream>
#include<algorithm>
using namespace std;
#define ls ftr[k].l
#define rs ftr[k].r
typedef double DB;
typedef long long LL;
typedef complex<double>CD;
const int maxn = 50010;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;

void Mod( int &x ){ x%=mod; }
int Min( int a , int b ){ return a<b?a:b; }
int Max( int a , int b ){ return a>b?a:b; }
int Random(){ static int seed = 703; return seed=(int)(seed*48271LL%2147483647); }

int sizes,root;

struct FHQ_Treap
{
    int l,r;
    int v,rnd;
    int sizes;
    int maxx,cnt,rev;
}ftr[maxn];

void Addone( int k , int s )
{
    if ( !k )
        return;
    ftr[k].v += s;
    ftr[k].cnt += s;
    ftr[k].maxx += s;
}

void Revone( int k )
{
    if ( !k )
        return;
    ftr[k].rev ^= 1;
    swap ( ls , rs );
}

void Push_up( int k )
{
    if ( !k )
        return;
    ftr[k].maxx = Max( ftr[k].v , Max ( ftr[ls].maxx , ftr[rs].maxx ) );
    ftr[k].sizes = ftr[ls].sizes+ftr[rs].sizes+1;
}

void Push_down( int k )
{
    if ( !k )
        return;
    if ( ftr[k].cnt )
        Addone( ls , ftr[k].cnt ),Addone( rs , ftr[k].cnt );
    if ( ftr[k].rev )
        Revone( ls ),Revone( rs );
    ftr[k].cnt = 0;
    ftr[k].rev = 0;
}

void Build( int &k , int s )
{
    k = ++sizes;
    ftr[k].l = 0;
    ftr[k].r = 0;
    ftr[k].v = s;
    ftr[k].rnd = Random();
    ftr[k].sizes = 1;
    ftr[k].maxx = s;
    ftr[k].cnt = 0;
    ftr[k].rev = 0;
}

void Split( int k , int &l , int &r , int s )
{
    if ( s==0 )
        l = 0,r = k;
    else if ( s==ftr[k].sizes )
        l = k,r = 0;
    else if ( s<=ftr[ls].sizes )
        Push_down( r=k ),Split( ls , l , ls , s ),Push_up( k );
    else
        Push_down( l=k ),Split( rs , rs , r , s-ftr[ls].sizes-1 ),Push_up( k );
}

void Merge( int &k , int l , int r )
{
    if ( l==0||r==0 )
        k = l+r;
    else if ( ftr[l].rnd<ftr[r].rnd )
        Push_down( k=l ),Merge( rs , rs , r ),Push_up( k );
    else
        Push_down( k=r ),Merge( ls , l , ls ),Push_up( k );
}

void Add( int l , int r , int s )
{
    int x,y,z;
    Split( root , x , y , r );
    Split( x , z , x , l-1 );
    Addone( x , s );
    Merge( x , z , x );
    Merge( root , x , y );
}

void Reverse( int l , int r )
{
    int x,y,z;
    Split( root , x , y , r );
    Split( x , z , x , l-1 );
    Revone( x );
    Merge( x , z , x );
    Merge( root , x , y );
}

int Query( int l , int r )
{
    int x,y,z,ans;
    Split( root , x , y , r );
    Split( x , z , x , l-1 );
    ans = ftr[x].maxx;
    Merge( x , z , x );
    Merge( root , x , y );
    return ans;
}

int main()
{
    for ( int n,m ; scanf( "%d%d" , &n ,&m )==2 ; )
    {
        ftr[0].maxx = -inf;
        sizes = root = 0;
        for ( int i=1,t ; i<=n ; i++ )
        {
            Build ( t , 0 );
            Merge ( root , root , t );
        }
        for ( int i=1 ; i<=m ; i++ )
        {
            int op,l,r,x;
            scanf ( "%d" , &op );
            if ( op==1 )
            {
                scanf ( "%d%d%d" , &l , &r , &x );
                Add( l , r , x );
            }
            else if ( op==2 )
            {
                scanf ( "%d%d" , &l , &r );
                Reverse ( l , r );
            }
            else
            {
                scanf ( "%d%d" , &l , &r );
                printf ( "%d\n" , Query( l , r ) );
            }
        }
    }
    return 0;
}

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转载自blog.csdn.net/Game_Acm/article/details/81359688