/*
Prototypes analyze
时间限制：1000 ms | 内存限制：65535 KB
难度：2
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
*/
题意：求不同形状的二叉树有多少个？
思路：首先创建二叉排序树，按照先序遍历二叉排序树，最后去掉相同的。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct Tree
{
Tree *l;
Tree *r;
int data;//存放输入的数据
int count;//存放编号
}*root,*p,*q;
int s[60][30];
int con,n,k;
Tree *c_Tree_root(int num,int x)
{
p = (struct Tree *)malloc(sizeof(Tree));
p->l=NULL;
p->r=NULL;
p->data = x;
p->count = num;
return p;
}
void Create_Tree(Tree *tree,int x)//创建二叉排序树
{
int num = tree->count;
if(x >= tree->data)//如果当前值大于树根的值，放树的右孩子上。
{
if(tree->r==NULL)//右孩子为NULL，放上面
{
q = c_Tree_root(num*2+1,x);
tree->r = q; tree = q;
return ;
}
else
{
Create_Tree(tree->r,x);//不为空，继续以右孩子为根查找
}
}
else
{
if(tree->l==NULL)
{
q = c_Tree_root(num*2,x);
tree->l=q; tree = q;
return ;
}
else
{
Create_Tree(tree->l,x);
}
}
}
void pre_Tree(Tree *tree,int i)//前序遍历 把编号存到数组中
{
if(tree!=NULL)
{
s[i][con++] = tree->count;
pre_Tree(tree->l,i);
pre_Tree(tree->r,i);
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int x,num;
scanf("%d%d",&n,&k);
int i,j,a;
memset(s,0,sizeof(s));
for(i=1;i<=n;i++)
{
scanf("%d",&x);
root = c_Tree_root(1,x);//先创建一个根节点
for(j=1;j<k;j++)
{
scanf("%d",&x);
Create_Tree(root,x); //创建二叉排序树
}
con = 1;
pre_Tree(root,i);
}
int tem;
for(i=1;i<n;i++)//去重
{
for(j=i+1;j<=n;j++)
{
for(a=1,tem=0;a<=k;a++)
{
if(s[i][a]==s[j][a] && (s[i][a]+s[j][a])!=0)
{
tem ++ ;
}
}
if(tem==k)
{
for(a=1;a<=k;a++)//找到相同的 把相同的标记为0
{
s[j][a] = 0;
}
}
}
}
int ans;
int res = n;
for(i=1;i<=n;i++)
{
for(j=1,ans=0;j<=k;j++)
{
if(s[i][j]==0)
{
ans ++ ;
}
}
if(ans==k)
{
res--;
}
}
printf("%d\n",res);
}
return 0;
}
