fzu 1962 树状数组 OR 线段树

优雅的树状数组！人们发明了复杂度为logn的求解第k小的方法，常数小且代码量小，实在是用来求解本类题目的最佳方法，可惜的是我的Treap超时了......

树状数组：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 500001;
int c[N];
int n, m, maxn;

int lb( int i )
{
    return i & -i;
}

void update( int i, int v )
{
    while ( i <= n )
    {
        c[i] += v;
        i += lb(i);
    }
}

int sum( int i )
{
    int ans = 0;
    while ( i )
    {
        ans += c[i];
        i -= lb(i);
    }
    return ans;
}

int kth( int k )
{
    int ans = 0, cnt = 0;
    for ( int i = 20; i >= 0; i-- )
    {
        ans += ( 1 << i );
        if ( ans >= maxn || cnt + c[ans] >= k )
        {
            ans -= ( 1 << i );
        }
        else
        {
            cnt += c[ans];
        }
    }
    return ans + 1;
}

int main ()
{
    while ( scanf("%d%d", &n, &m) != EOF )
    {
        memset( c, 0, sizeof(c) );
        for ( int i = 1; i <= n; i++ )
        {
            update( i, 1 );
        }
        maxn = n;
        while ( m-- )
        {
            char op[2];
            int num;
            scanf("%s%d", op, &num);
            if ( op[0] == 'L' )
            {
                int kk = kth(num);
                update( kk, -1 );
            }
            else if ( op[0] == 'R' )
            {
                if ( sum(num) == sum( num - 1 ) )
                {
                    update( num, 1 );
                }
            }
            else if ( op[0] == 'Q' )
            {
                int kk = kth(num);
                printf("%d
", kk);
            }
        }
    }
    return 0;
}

线段树当然也可以做：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 500001;
int n, m;

struct Node 
{
    int l, r, sum;
} node[N << 2];

void pushup( int i )
{
    node[i].sum = node[i << 1].sum + node[i << 1 | 1].sum;
}

void build( int i, int l, int r )
{
    node[i].l = l, node[i].r = r, node[i].sum = node[i].r - node[i].l + 1;
    if ( l == r ) return ;
    int mid = ( l + r ) >> 1;    
    build( i << 1, l, mid );
    build( i << 1 | 1, mid + 1, r );
}

void update( int i, int pos, int val )
{
    if ( node[i].l == pos && node[i].r == pos )
    {
        node[i].sum = val;
        return ;
    }
    int mid = ( node[i].l + node[i].r ) >> 1;
    if ( pos <= mid )
    {
        update( i << 1, pos, val );
    }
    else
    {
        update( i << 1 | 1, pos, val );
    }
    pushup(i);
}

int query( int i, int v )
{
    if ( node[i].l == node[i].r ) return node[i].l;
    if ( node[i << 1].sum >= v ) return query( i << 1, v );
    return query( i << 1 | 1, v - node[i << 1].sum );
}

int main ()
{
    while ( scanf("%d%d", &n, &m) != EOF )
    {
        build( 1, 1, n );
        scanf("%d", &m);
        while ( m-- )
        {
            char op[2];
            int x;
            scanf("%s %d", op, &x);
            if ( op[0] == 'L' )
            {
                int kk = query( 1, x );
                update( 1, kk, 0 );
            }
            else if ( op[0] == 'R' )
            {
                update( 1, x, 1 );
            }
            else if ( op[0] == 'Q' )
            {
                int kk = query( 1, x );
                printf("%d
", kk);
            }
        }
    }
    return 0;
}