Backtracking [Guide]

#	User	Rating
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

#	User	Contrib.
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	159
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	Dominater069	153
9	nor	153

Backtracking: This algorithm helps you solve problems in a specific way using your skills in recursive programming. It comes in handy whenever you have a set of valid answers to a problem and need to search through them all or construct a suitable answer, this algorithm is a general algorithm for finding all or some solutions in some computational problems. The backtracking method is an algorithmic method to solve some kind of problems in which we need to build and examine all possible situations or a number of good situations of an arrangement(combination), in this algorithm all possible situations are created , and after examining each one, if it is approved based on the request of the problem we will accept it as a answer(kind of like brute force we find each possible combination). Note that this method is useful when we cant use a faster algorithm(like: Greedy or Dp).

the General Backtracking Method is like this(pseudocode):

solve(X):
if X is accepted:
   X is answer
   exit
while we can change X:
   change X
   if X is valid:
      solve(X)
   undo last change from X

The base of the algorithm is in 4 parts: the condition of finding the answer (lines 1 to 3) — finding all the different states of the answer (lines 4 to 6) — recursive call (line 7) — return the last change made (line 8).

Some Backtracking Problems:

1-Print all strings of length n that are made only of letters a, b, and c.

The Answer

Backtracking method: We perform the same general method, we add one of the letters to the string each round and print it when the length of the string is equal to n, and until the length of the string is less than is n, we repeat this process (if the length of the string is greater than n, we backtrack), code in C++

vector<char> temp;
void print(){
   for (int i = 0; i < n; i++) cout << temp[i];
   cout << "\n";
}
void solve(int i){
   if (i == n) print();
   else if (i > n) return;
   else {
      char chr[3] = {‘a’ , ‘b’ , ‘c’};
      for (int j = 0; j < 3; j++){
         //changing
         temp.push_back(chr[j]);
         //recursive call
         solve(i+1);
         //undo last change
         temp.pop_back();
      }
   }
}

2-A horse is in the cell (0,0) in a n*n chessboard. That is, in the upper-left corner, you have to say the number of cells that the horse can reach with exactly k moves.

The Answer

Backtracking method: we move till, when the number of moves is equal to k, if a new cells found, we add it to our answers (because we may reach the same cells with two different methods). otherwise, if the moves made were more than k, we go back (we backtrack), and if it was less than k, we move the 8 possible moves of the horse and with the condition that after the move we stay inside the chess board, we do each movement again and... , code in C++(result is saved in ans variable)

int ans;
vector<int> answers;
bool inside(int x , int y){
   return x >= 0 && x < n && y >= 0 && y < n;
}
void solve(int x , int y , int i){
   if(i == k && !count(answers.begin() , answers.end() , make_pair(x , y)){
      answers.push_back(make_pair(x , y));
      ans++;
      return;
   }else if(i > k) return;
   else{
      //all possible moves by a knight
      int dx[8] = {1 , 2 , -1 , -2 , -1 , -2 , 1 , 2};
      int dy[8] = {2 , -1 , -2 , 1 , 2 , -1 , -2 , 1};
      for(int j = 0 ; j < 8 ; j++){
         int new_x = x + dx[j];
         int new_y = y + dy[j];
         //changing
         if(inside(new_x , new_y)) //checking is valid
            solve(new_x , new_y , i+1); //recursive call
      }
   }
}

3-476B - Dreamoon and WiFi

The Answer

Backtracking method: in the string each time we reach a '?' , first we will turn it into '-' then into '+' (or reversed) , and when there is no more '?' we count, if it is accepted we will increase the good ways.

#include <bits/stdc++.h>
using namespace std;

string s1;
string s2;
double goods= 0;
double all = 0;
 
void backtrack(int i , int res){
	if(s2[i] == '?'){
                //changing
		s2[i] = '-';
                //recursive call
		backtrack(i , res);
		s2[i] = '+';
		backtrack(i , res);
                //undo last change
		s2[i] = '?';	
	}else if(i == s1.length()-1){
		int ans = 0;
		all++;
		for(auto c : s2){
			if(c == '+') ans++;
			else ans--;	
		}
		if(ans == res) goods++;
 
		return;
	}
	if(i >= s1.length()){
		return;
	}
	if(s2[i] == '+'){ 
		backtrack(i+1 , res);
	}
	else if(s2[i] == '-'){
		backtrack(i+1 , res);
	}
	return;
}

int main() {
    ios::sycn_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> s1 >> s2;
    int res = 0;
    for(auto c : s1){
    	if(c == '+') res++;
	else res--;
    }
    backtrack(0 , res);
    cout << setprecision(9) << goods/all;
}

AlRntn's blog