hdu 3308 线段树＋区间合并

最新推荐文章于 2021-09-06 21:36:41 发布

黎辰

最新推荐文章于 2021-09-06 21:36:41 发布

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文章标签： C++ 线段树区间合并

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/44039307

本文探讨了区间合并问题的解决方法，通过维护左最大、最大和右最大值，实现高效的数据结构优化。通过实例代码展示了解决过程，并在不同场景下进行了应用演示。

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很明显是区间合并的题，只需要维护左最大，最大，和右最大即可，在合并时需要讨论．

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MAX 100007

using namespace std;

int t,n,m,u,v,a[MAX];
char s[5];

struct Tree
{
    int l,r,left_mx , mx , right_mx;
}tree[MAX<<2];

void push_up ( int u )
{
    tree[u].left_mx = tree[u<<1].left_mx;
    tree[u].right_mx = tree[u<<1|1].right_mx;
    tree[u].mx = max ( tree[u<<1].mx , tree[u<<1|1].mx );
    if ( a[tree[u<<1].r] >= a[tree[u<<1|1].l] ) return;
    if ( tree[u<<1].left_mx == tree[u<<1].r - tree[u<<1].l + 1 )
        tree[u].left_mx = tree[u<<1].left_mx + tree[u<<1|1].left_mx;
    if ( tree[u<<1|1].right_mx == tree[u<<1|1].r - tree[u<<1|1].l + 1 )
        tree[u].right_mx = tree[u<<1].right_mx + tree[u<<1|1].right_mx;
    tree[u].mx = max ( tree[u].mx , tree[u<<1].right_mx + tree[u<<1|1].left_mx );
}

void set ( int u )
{
    tree[u].left_mx = tree[u].right_mx 
                    = tree[u].mx = 1;
}

void build ( int u , int l , int r )
{
    tree[u].l = l , tree[u].r = r;
    if ( l == r ) 
    {
        set ( u );
        return;
    }
    int mid = l + r >> 1;
    build ( u<<1 , l , mid );
    build ( u<<1|1 , mid+1 , r );
    push_up ( u );
}

void update ( int u , int x , int v )
{
    int l = tree[u].l , r = tree[u].r;
    if ( l == r )
    {
        a[x] = v;
        return;
    }
    int mid = l + r >> 1;
    if ( x > mid ) update ( u<<1|1 , x , v );
    else update ( u<<1 , x , v );
    push_up ( u );
}

struct Ret
{
    int l,r,mx,l_mx,r_mx;
    Ret ( int a , int b , int c , int d , int e )
        :l(a),r(b),mx(c),l_mx(d),r_mx(e){}
    Ret ( ) { }
};

Ret query ( int u , int left , int right )
{
    int l = tree[u].l , r = tree[u].r;
    if ( left <= l && r <= right )
        return Ret ( tree[u].l , tree[u].r , tree[u].mx,
                     tree[u].left_mx , tree[u].right_mx );
    int mid = l + r >> 1;
    Ret ret1 , ret2 , ret;
    int flag = 0;
    if ( left <= mid && right >= l )
       ret1 = query( u<<1 , left , right ),flag++;
    if ( left <= r && right > mid )
       ret2 = query ( u<<1|1 , left , right ),flag += 2;
    if ( flag == 1 ) return ret1;
    else if ( flag == 2 ) return ret2;
    else
    {
        ret.l = ret1.l , ret.r = ret2.r;
        ret.mx = max ( ret1.mx , ret2.mx );
        ret.l_mx = ret1.l_mx;
        ret.r_mx = ret2.r_mx;
        if ( a[ret1.r] >= a[ret2.l] )
            return ret;
        if ( ret1.l_mx == ret1.r - ret1.l + 1 )
            ret.l_mx = ret1.l_mx + ret2.l_mx;
        if ( ret2.r_mx == ret2.r - ret2.l + 1 )
            ret.r_mx = ret1.r_mx + ret2.r_mx;
        ret.mx = max ( ret.mx , ret1.r_mx + ret2.l_mx );
        return ret;
    }
     
}

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%d%d" , &n , &m );
        for ( int i = 1 ; i <= n ; i++ )
            scanf ( "%d" , &a[i] );
        build ( 1 , 1 , n );
        for ( int i = 0 ; i < m ; i++ )
        {
            scanf ( "%s" , s );
            scanf ( "%d%d" , &u , &v );
            u++;
            if ( s[0] == 'U' ) 
            {
                update ( 1 , u , v );
                /*for ( int i = 1 ; i <= n ; i++ )
                    printf ( "%d " , a[i] );
                puts ( "" );*/
            }
            else 
            {
                v++;
                printf ( "%d\n" , query ( 1 , u , v ).mx ); 
            }
        }
    }
}