題目：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.
題目大意：在一個無環多叉樹之中，從根節點出發用深度優先遍歷去遍歷每個元素，遠離根節點用0表示；靠近根節點用1表示。現在有兩個字符串，看其能否是同一顆多叉樹的深度優先遍歷。
思路：一棵樹相同當且僅當其子樹全相同；判斷起來有點複雜。則弱化一下，當且僅當這兩棵樹的所有子樹的元素個數相等，而且個數相等的子樹在原樹中的深度相等。例如下圖：（弱化的當且僅當不知道正確與否！！沒檢查過是完備）
| 節點 | 除根節點子樹元素個數（及根深度） |
| 1 | 5（1） |
| 2 | 2（2） |
| 3 | 0（2） |
| 4 | 0（2） |
| 5 | 0（3） |
| 6 | 0（3） |
//284K 32MS
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct obj{
short int len;
short int depth;
obj():len(0),depth(0){};
};
bool judge(string s1,string s2);//判斷s1、s2是否爲同一棵樹的深度優先遍歷
void transfor(string samp, vector<obj>& ret);//把str變成上面講的所有子樹原數個數和深度的結構數組
bool test_equ(vector<obj>& obj1,vector<obj>& obj2);//判斷兩個排序後的數組是否爲完全相同
bool cmp(const obj &a ,const obj &b)//重載比較函數，因爲要對結構體排序
{
if(a.len < b.len)
return true;
if(a.len==b.len && a.depth<b.depth)
return true;
return false;
}
int main()
{
const string result[2]={"different","same"};
int num;
string s1,s2;
vector<obj> s12int,s22int;
cin>>num;
while(num--){
s1.clear();s2.clear();
s12int.clear();s22int.clear();
cin>>s1>>s2;
cout<<result[judge(s1,s2)]<<endl;
}
// system("pause");
return 1;
}
bool judge(string s1,string s2){
if(s1.size() != s2.size())
return false;
vector<obj> s12int,s22int;
s12int = vector<obj>(s1.size()); transfor(s1,s12int);
s22int = vector<obj>(s2.size()); transfor(s2,s22int);
sort(s12int.begin(),s12int.end(),cmp);
sort(s22int.begin(),s22int.end(),cmp);
return test_equ(s12int,s22int);
}
void transfor(string samp, vector<obj>& ret){
int len = samp.size(),sum,dep=0;
if(len != ret.size())
return;
for(int i=0;i<len;i++){
if(samp[i] == '1'){
dep--;
ret[i].depth = dep;
}else{
dep++;
ret[i].depth = dep;
}
}
for(int i=0;i<len;i++){
if(samp[i]=='1'){
ret[i].len = 0;
continue;
}
sum=0;
for(int j=i;j<len;j++){
if(samp[j]=='1'){
sum++;
}else{
sum--;
}
if(sum==0){
ret[i].len = j-i+1;//可以優化，前面判斷一些ret[]之後可以知道某些ret[]可以不計算
break;
}
}//for(j)
}//for(i)
}//end of fun
bool obj_equ(obj t1,obj t2){
if((t1.depth == t2.depth) && (t1.len == t2.len))
return true;
return false;
}
bool test_equ(vector<obj>& obj1,vector<obj>& obj2){
int len = obj1.size();
for(int i=0;i<len;i++){
if(obj_equ(obj1[i],obj2[i]) == false)
return false;
}
return true;
}遇到網上搜那個有些代碼只判斷子樹元素個數，，沒判斷樹深。會遇到下面的例子判斷錯誤結果！！！但是依然能AC？！但是也不知道我的代碼有沒有BUG。。。
0010010111000111
0010001111001011
自己想了一下還是有問題，在處理這個示例的時候：
0010001011110000011111
0010000111110000101111
又在網上找了一下：發現有人用樹的最小表示方法做。不是動態規劃。就是遞歸將樹的子樹排序，然後整棵樹就變成了一個有序樹，然後再比較字符串就可以了。代碼如下：（轉至http://blog.csdn.net/u012433233/article/details/24424789）
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
string min_pre(string str){
vector<string> box;
string ret = "";
int equal = 0, st = 0;
for(int i = 0; i < str.size(); i++){
if(str[i] == '0')
equal++;
else
equal--;
if(equal == 0){
if(i - 1 > st + 1){
box.push_back("0" + min_pre(str.substr(st + 1,i - 1 - st)) + "1");
}else
box.push_back("01");
st = i + 1;
}
}
sort(box.begin(), box.end());
for(int i = 0; i < box.size(); i++)
ret += box[i];
return ret;
}
int main() {
int t;
cin >> t;
while (t--) {
string tree1, tree2;
cin >> tree1 >> tree2;
if (min_pre(tree1) == min_pre(tree2))
cout << "same" << endl;
else
cout << "different" << endl;
}
return 0;
}