nyoj 1278 Prototypes analyze（平衡二叉树）

Prototypes analyze

描述
ALpha Ceiling Manufacturers (ACM) is analyzing the properties of its new series of Incredibly Collapse-Proof Ceilings (ICPCs). An ICPC consists of n layers of material, each with a different value of collapse resistance (measured as a positive integer). The analysis ACM wants to run will take the collapse-resistance values of the layers, store them in a binary search tree, and check whether the shape of this tree in any way correlates with the quality of the whole construction. Because, well, why should it not? To be precise, ACM takes the collapse-resistance values for the layers, ordered from the top layer to the bottom layer, and inserts them one-by-one into a tree. The rules for inserting a value v are:

• If the tree is empty, make v the root of the tree.

• If the tree is not empty, compare v with the root of the tree.

• If v is smaller, insert v into the left subtree of the root,

• otherwise insert v into the right subtree.

ACM has a set of ceiling prototypes it wants to analyze by trying to collapse them. It wants to take each group of ceiling prototypes that have trees of the same shape and analyze them together. For example , assume ACM is considering five ceiling prototypes with three layers each, as described by Sample Input 1 and shown in Figure C.1. Notice that the first prototype’s top layer has collapseresistance value 2, the middle layer has value 7, and the bottom layer has value 1. The second prototype has layers with collapse-resistance values of 3, 1, and 4 – and yet these two prototypes induce the same tree shape, so ACM will analyze them together. Given a set of prototypes, your task is to determine how many different tree shapes they induce.

输入
The first line of the input contains one integers T, which is the nember of test cases (1<=T<=8).
Each test case specifies :
● Line 1: two integers n (1 ≤ n ≤ 50), which is the number of ceiling prototypes to analyze,
and k (1 ≤ k ≤ 20), which is the number of layers in each of the prototypes.
● The next n lines describe the ceiling prototypes. Each of these lines contains k distinct
integers ( between 1 and 1e6, inclusive ) , which are the collapse-resistance values of the
layers in a ceiling prototype, ordered from top to bottom.
输出
For each test case generate a single line containing a single integer that is the number of different tree
shapes.
样例输入
1
5 3
2 7 1
1 5 9
3 1 4
2 6 5
9 7 3
样例输出
4
来源
河南省第九届省赛


参考：二叉排序树

题意：让你建n个二叉排序树，问最后有多少种形状不同的树

思路：先建树，然后直接暴力，通过先序遍历比较树即可

代码：

#include<stdio.h>typedef struct node{    int key;    struct node *leChild,*riChild;}Node,*BST;int flag;void BSTInsert(BST &p,int element)//建树{    if(NULL==p)    {        p=new Node;        p->key=element;        p->leChild=p->riChild=NULL;        return ;    }    if(element<(p->key))        BSTInsert(p->leChild,element);    else        BSTInsert(p->riChild,element);}void judge(BST t1,BST t2)//比较树的形状{    if(t1==NULL&&t2==NULL)        return ;    else if(t1&&t2)    {        judge(t1->leChild,t2->leChild);        judge(t1->riChild,t2->riChild);    }    else flag=0;}int main(){    int t,n,k,x;    BST tree[55];    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        for(int i=0; i<n; ++i)        {            BST T=NULL;            for(int j=0; j<k; ++j)            {                scanf("%d",&x);                BSTInsert(T,x);            }            tree[i]=T;        }        int ans=0;        for(int i=0;i<n;++i)        {            int flog=1;            for(int j=i+1;j<n;++j)            {                flag=1;                judge(tree[i],tree[j]);                if(flag)                {                    flog=0;                    break;                }            }            if(flog)                ++ans;        }        printf("%d\n",ans);    }    return 0;}