POJ 1635 Subway tree systems 有根树的同构

POJ 1635

题目很简单 给个3000节点以内的根确定的树 判断是否同构。用Hash解决，类似图的同构，不过效率更高。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<cmath>
#include<vector>
using namespace std;

typedef unsigned long long int ULL;

const ULL leaf_hash=2099,pt=9907099,qt=9907099;
const int maxn=3000+1;
char str1[maxn],str2[maxn] ;
char *p;

ULL Hash()
{
 ULL sum=1;
 if(*(p)=='1'&&*(p-1)=='0')
 	{
 	 p++;
 	 return leaf_hash;
	}
 while(*p!='\0'&&*p++=='0')//这个巧妙的循环,把子节点的hash值都加给了父节点,作为父节点的hash值
     {
     	sum=(sum*(pt^Hash()))%qt;
	 }
 return sum;
}
int main()
{
 int T;scanf("%d",&T);
 while(T--)
 	{
 	 scanf("%s%s",str1,str2);
 	 p=str1;
 	 ULL a=Hash();
 	 p=str2;
 	 ULL b=Hash();
 	 if(a==b)puts("same");
			else  puts("different");
	}
 return 0;
}