poj 1635

有根树同构。参考论文《hash在。。。。》

#include <iostream>
#include <fstream>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cmath>

using namespace std;

const int leaf_hash=2099;
const int pt=9323;
const int qt=8719;
char str1[3100], str2[3100];
char *p;

int Hash()
{
    int sum=1;
    if(*(p)=='1'&&*(p-1)=='0'){
        p++;
        //cout<<"leaf_hash="<<endl;
        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
    {
        //cout<<"for"<<endl;
        sum = (sum*(pt^Hash()))%qt;
    }
//    printf("sum==%d\n",sum);
    return sum;
}


int main()
{
//    freopen("input.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s%s", str1, str2);
        p = str1;
        int a = Hash();
        p = str2;
        //cout<<a<<endl;
        int b = Hash();
        //cout<<b<<endl;
        if(a == b)
        {
            puts("same");
        }
        else
        {
            puts("different");
        }
    }
    return 0;
}