POJ 2985 The kth Largest Group

POJ_2985

    这个题目以cats的个数为关键字建立线段树去做，我为了练一下昨天刚学的SBT所以就用SBT写了，不过SBT的效率要低一些。

#include<stdio.h>
#include<string.h>
#define MAXD 200010
#define MAXM 400010
int N, M, T, node, p[MAXD], left[MAXM], right[MAXM], key[MAXM], size[MAXM], s[MAXM];
int find(int x)
{
    return p[x] == x ? x : (p[x] = find(p[x]));
}
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 v)
{
    T = ++ node;
    size[T] = 1;
    key[T] = v;
    left[T] = right[T] = 0;
}
void Insert(int &T, int v)
{
    if(T == 0)
    {
        add(T, v);
        return ;
    }
    ++ size[T];
    if(v < key[T])
        Insert(left[T], v);
    else
        Insert(right[T], v);
    maintain(T, v >= key[T]);
}
int Delete(int &T, int v)
{
    -- size[T];
    if(key[T] == v || (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;
    }
    else if(v < key[T])
        return Delete(left[T], v);
    else
        return Delete(right[T], v);
}
int select(int &T, int k)
{
    int n = size[left[T]] + 1;
    if(k == n)
        return key[T];
    if(k < n)
        return select(left[T], k);
    else
        return select(right[T], k - n);
}
void init()
{
    int i;
    T = node = left[0] = right[0] = size[0] = 0;
    for(i = 1; i <= N; i ++)
    {
        p[i] = i, s[i] = 1;
        Insert(T, 1);
    }
}
void solve()
{
    int i, j, k, x, y, tx, ty;
    for(i = 0; i < M; i ++)
    {
        scanf("%d", &j);
        if(j == 0)
        {
            scanf("%d%d", &x, &y);
            tx = find(x), ty = find(y);
            if(tx != ty)
            {
                Delete(T, s[tx]);
                Delete(T, s[ty]);
                p[ty] = tx;
                s[tx] += s[ty];
                Insert(T, s[tx]);
            }
        }
        else
        {
            scanf("%d", &k);
            printf("%d\n", select(T, size[T] - k + 1));
        }
    }
}
int main()
{
    while(scanf("%d%d", &N, &M) == 2)
    {
        init();
        solve();
    }
    return 0;
}