## poj-1635 Subway tree systems（判断两个有根树是否同构）-哈希法

2017年05月14日 ⁄ 综合 ⁄ 共 3487字 ⁄ 字号 评论关闭

Description

Some major cities have subway systems in the form of a tree, i.e. between any pair of stations, there is one and only one way of going by subway. Moreover, most of these cities have a unique central station. Imagine you are a tourist in one of these cities
and you want to explore all of the subway system. You start at the central station and pick a subway line at random and jump aboard the subway car. Every time you arrive at a station, you pick one of the subway lines you have not yet travelled on. If there
is none left to explore at your current station, you take the subway line back on which you first came to the station, until you eventually have travelled along all of the lines twice,once for each direction. At that point you are back at the central station.
Afterwards, all you remember of the order of your exploration is whether you went further away from the central station or back towards it at any given time, i.e. you could encode your tour as a binary string, where 0 encodes taking a subway line getting you
one station further away from the central station, and 1 encodes getting you one station closer to the central station.

Input

On the first line of input is a single positive integer n, telling the number of test scenarios to follow.Each test scenario consists of two lines, each containing a string of the characters '0' and '1' of length at most 3000, both describing a correct exploration
tour of a subway tree system.

Output

exploration tours of the same subway tree system, or the text "different" if the two strings cannot be exploration tours of the same subway tree system.

Sample Input

```2
0010011101001011
0100011011001011
0100101100100111
0011000111010101```

Sample Output

```same
different```

```import java.util.*;
public class SubwayTreeSstems1635 {

static final int Hn=11000;
static int h[]=new int[Hn];
static Random rand=new Random(System.currentTimeMillis());
static int m=1000000007;
static int index=0;
/**
* @param args
*/
public static void main(String[] args) {

run();
}

private static void init() {

for(int i=0;i<Hn;i++)
h[i]=(rand.nextInt()%m);
}

public static void run()
{
Scanner in=new Scanner(System.in);
int T=in.nextInt();
init();
for(int t=0;t<T;t++)
{
String s1=in.next();
Node tree1=createTree(s1);
String s2=in.next();
Node tree2=createTree(s2);
/*System.out.println(tree1.children.size()+" "+tree2.children.size());
displayTree(tree1);
System.out.println();
displayTree(tree2);*/

int a=hash(tree1,1);
int b=hash(tree2,1);
//System.out.println(a+" "+b);
if(a==b)
{
System.out.println("same");
}
else
{
System.out.println("different");
}
}
}

public static int hash(Node tree,int j)
{
int sum=h[j+5000];//j是树的高度
for(Node n:tree.children)
sum=(sum+h[j]*hash(n,j+1))%m;//把子树的哈希值加到父节点上去
return (sum*sum)%m;

}

private static Node createTree(String s) {

char[] seq=s.toCharArray();
Node root=new Node(0);
Node p=root;
int index=1;
for(int i=0;i<seq.length;i++)
{
if(seq[i]=='0')
{
Node node =new Node(index++);
connect(p,node);
p=node;
}
else if(seq[i]=='1')
{
p=p.parent;
}
}
//if(p==root)
//	System.out.println("create success!");
return root;
}

private static void connect(Node p, Node node) {

node.parent=p;
}

public static void displayTree(Node tree)
{
System.out.println(tree);
for(Node ch:tree.children)
displayTree(ch);
}

}

class Node
{
int id;
Node parent=null;
List<Node> children=new ArrayList<Node>();
public Node(int n)
{
id=n;
}
public String toString()
{
StringBuilder sb=new StringBuilder();
sb.append(id).append(": ");
for(Node n:children)
sb.append(n.id).append(" ");
return sb.toString();
}
}
```