POJ 3481 Double Queue

POJ_3481

    为了练一下昨天刚学的SBT，所以就用SBT写了，但实际上这个题目可以通过维护一个最大堆和一个最小堆就能实现所有操作。

#include<stdio.h>
#include<string.h>
#define MAXD 1000010
int T, node, key[MAXD], client[MAXD], left[MAXD], right[MAXD], size[MAXD];
void left_rotate(int &T)
{
    int k = right[T];
    right[T] = left[k];
    left[k] = T;
    size[k] = size[T];
    size[T] = size[left[T]] + size[right[T]] + 1;
    T = k;
}
void right_rotate(int &T)
{
    int k = left[T];
    left[T] = right[k];
    right[k] = T;
    size[k] = size[T];
    size[T] = size[left[T]] + size[right[T]] + 1;
    T = k;
}
void maintain(int &T, int flag)
{
    if(flag == 0)
    {
        if(size[left[left[T]]] > size[right[T]])
            right_rotate(T);
        else if(size[right[left[T]]] > size[right[T]])
            left_rotate(left[T]), right_rotate(T);
        else
            return ;
    }
    else
    {
        if(size[right[right[T]]] > size[left[T]])
            left_rotate(T);
        else if(size[left[right[T]]] > size[left[T]])
            right_rotate(right[T]), left_rotate(T);
        else
            return ;

    }
    maintain(left[T], 0);
    maintain(right[T], 1);
    maintain(T, 0);
    maintain(T, 1);
}
void add(int &T, int c, int v)
{
    T = ++ node;
    size[T] = 1;
    client[T] = c;
    key[T] = v;
    left[T] = right[T] = 0;
}
void Insert(int &T, int c, int v)
{
    if(T == 0)
    {
        add(T, c, v);
        return ;
    }
    ++ size[T];
    if(v < key[T])
        Insert(left[T], c, v);
    else
        Insert(right[T], c, v);
    maintain(T, v >= key[T]);
}
int Delete(int &T, int v)
{
    if(v == key[T] || (v < key[T] && left[T] == 0) ||(v > key[T] && right[T] == 0))
    {
        int k = key[T];
        if(left[T] == 0 || right[T] == 0)
            T = left[T] + right[T];
        else
            key[T] = Delete(left[T], key[T] + 1);
        return k;
    }
    if(v < key[T])
        return Delete(left[T], v);
    else
        return Delete(right[T], v);
}
void select(int &T, int k, int &c, int &v)
{
    int n = size[left[T]] + 1;
    if(n == k)
    {
        c = client[T];
        v = key[T];
        Delete(T, v);
        return ;
    }
    -- size[T];
    if(k < n)
        select(left[T], k, c, v);
    else
        select(right[T], k - n, c, v);
}
int main()
{
    int k, c, v;
    node = T = left[0] = right[0] = size[0] = 0;
    for(;;)
    {
        scanf("%d", &k);
        if(!k)
            break;
        if(k == 1)
        {
            scanf("%d%d", &c, &v);
            Insert(T, c, v);
        }
        else if(k == 2)
        {
            if(size[T])
            {
                select(T, size[T], c, v);
                printf("%d\n", c);
            }
            else
                printf("0\n");
        }
        else
        {
            if(size[T])
            {
                select(T, 1, c, v);
                printf("%d\n", c);
            }
            else
                printf("0\n");
        }
    }
    return 0;
}