SBT poj3481

Size Balanced Tree（SBT）平衡二叉树

http://blog.youkuaiyun.com/acceptedxukai/article/details/6921334

例题：poj3481

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string.h>
#include <queue>
#define N 100010
using namespace std;  //poj3481
//SBT模板题
struct SBT
{
    int key,left,right,size,num;     //num是特别添加的（题意）
} tree[N];

int root,top;  

void left_rot(int &x)
{
    int y = tree[x].right;
    tree[x].right = tree[y].left;
    tree[y].left = x;
    tree[y].size = tree[x].size;//转上去的节点数量为先前此处节点的size
    tree[x].size = tree[tree[x].left].size + tree[tree[x].right].size + 1;
    x = y;
}

void right_rot(int &x)
{
    int y = tree[x].left;
    tree[x].left = tree[y].right;
    tree[y].right = x;
    tree[y].size = tree[x].size;
    tree[x].size = tree[tree[x].left].size + tree[tree[x].right].size + 1;
    x = y;
}

void maintain(int &x,bool flag)
{
    if(flag == false)//左边
    {
        if(tree[tree[tree[x].left].left].size > tree[tree[x].right].size)//左孩子的左子树大于右孩子
            right_rot(x);
        else if(tree[tree[tree[x].left].right].size > tree[tree[x].right].size)//右孩子的右子树大于右孩子
        {
            left_rot(tree[x].left);
            right_rot(x);
        }
        else return;
    }
    else //右边
    {
        if(tree[tree[tree[x].right].right].size > tree[tree[x].left].size)//右孩子的右子树大于左孩子
            left_rot(x);
        else if(tree[tree[tree[x].right].left].size > tree[tree[x].left].size)//右孩子的左子树大于左孩子
        {
            right_rot(tree[x].right);
            left_rot(x);
        }
        else return;
    }
    maintain(tree[x].left,false);
    maintain(tree[x].right,true);
    maintain(x,true);
    maintain(x,false);
}

/*
*insert没有合并相同的元素，如果出现相同的元素则把它放到右子树上，这样能保证求第k小数的时候对相同元素也能正确
*/
void insert(int &x,int key, int num)
{
    if(x == 0)
    {
        x = ++top;
        tree[x].left = tree[x].right = 0;
        tree[x].size = 1;
        tree[x].key = key;
		tree[x].num=num;
    }
    else
    {
        tree[x].size ++;
        if(key < tree[x].key) insert(tree[x].left,key,num);
        else  insert(tree[x].right,key, num);//相同元素插入到右子树中
        maintain(x, key >= tree[x].key);//每次插入把平衡操作压入栈中
    }
}
int  remove(int &x,int key)
{
    int d_key;
    //if(!x) return 0;
    tree[x].size --;
    if((key == tree[x].key)||(key < tree[x].key && tree[x].left == 0) ||
            (key>tree[x].key && tree[x].right == 0))
    {
        d_key = tree[x].key;
        if(tree[x].left && tree[x].right)
        {
            tree[x].key = remove(tree[x].left,tree[x].key+1);
        }
        else
        {
            x = tree[x].left + tree[x].right;
        }
    }
    else if(key > tree[x].key)
        d_key = remove(tree[x].right,key);
    else if(key < tree[x].key)
        d_key = remove(tree[x].left,key);
    return d_key;
}
int getmin()
{
    int x, k;
    for(x = root ; tree[x].left; x = tree[x].left);
    k=tree[x].num;
	remove(root, tree[x].key); //输出最小值后要删除这个节点
	return k;
}

int getmax()
{
    int x, k;
    for(x = root ; tree[x].right; x = tree[x].right);
    k=tree[x].num;
	remove(root, tree[x].key);   //输出最大值后要删除这个节点
	return k;
}

int main()
{
	int n, a, b;
	root=top=0;
	while(scanf("%d", &n)&&n)
	{
		if(n==1)
		{
			scanf("%d%d" ,&a, &b);
			insert(root, b, a);
		}
		if(n==2)
		{
			printf("%d\n", getmax());
		}
		if(n==3)
		{
			printf("%d\n", getmin());
		}
	}
	return 0;
}