POJ1635（Subway tree systems）

Subway tree systems

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7161 Accepted: 2982

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

20010011101001011010001101100101101001011001001110011000111010101

Sample Output

samedifferent

题意：给出两棵树的深搜遍历表示，0表示深度加1，1表示回溯，深度减1。问两棵树是否相同。

思路：

方法一：dfs

#include<cstdio>#include<vector>#include<algorithm>#include<string>#include<iostream>#include<cstring>using namespace std;char a[3005],b[3005];string dfs(char *arr, int st, int ed){    if(st>=ed){        return "01";    }    int cnt=0,j=st;    vector<string> re;    re.clear();    for(int i = st; i <= ed; i++){        if(arr[i]=='0')cnt++;        else cnt--;        if(cnt==0){            re.push_back(dfs(arr,j+1,i-1));            j=i+1;        }    }    sort(re.begin(),re.end());    string ans="0";    for(int i = 0; i < re.size();i++)        ans+=re[i];    ans+="1";    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%s%s",a,b);        int len1=strlen(a),len2=strlen(b);        if(len1!=len2){            printf("different\n");            continue;        }        string ans1=dfs(a,0,len1-1);        string ans2=dfs(b,0,len2-1);        if(ans1==ans2)printf("same\n");        else printf("different\n");    }    return 0;}

方法二：

hash大法。

#include<cstdio>#include<cstdlib>#include<iostream>#include<cstring>using namespace std;const int Mod=3001,maxn=3005;char a[maxn],b[maxn];int has[maxn];char *now;int _hash(int cur){    int re=has[cur];    while(*now!='\0'&&*now++=='0'){        re = (re+_hash(cur+1)*has[cur])%Mod;    }    return (re*re)%Mod;}int main(){    int t;    for(int i = 0; i < maxn; i++)        has[i]=rand()%Mod;    scanf("%d",&t);    while(t--){        scanf("%s%s",a,b);        int len1=strlen(a),len2=strlen(b);        if(len1!=len2){            printf("different\n");            continue;        }        now=a;        int ans1=_hash(1);        now=b;        int ans2=_hash(1);        if(ans1==ans2)printf("same\n");        else printf("different\n");    }    return 0;}