tourist in the last contest has achieved 1st place and crossed 4000 rating and become the first person on codeforces in the rank Tourist

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.

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.

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

#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;
}

Here are Some easy to intermediate codeforces graph and backtracking problems this problems normally only use some basic knowledge of graphs/trees + basic algorithms , I hope it helps

1944A - Destroying Bridges rate : 800 -> basic of graphs [Very Simple]

1598A - Computer Game rate : 800 -> not so much graph [Very Simple]

1797A - Li Hua and Maze rate : 800 -> not so much graph + edge cases [Very Simple]

1843C - Sum in Binary Tree rate : 800 -> not so much graph or tree [Very Simple]

939A - Love Triangle rate : 800 -> can be solved with graph or not [Very Simple]

115A - Party and 116C - Party rate : 900 -> bfs [Simple]

500A - New Year Transportation rate : 1000 -> dfs [Simple]

727A - Transformation: from A to B rate : 1000 -> backtracking / dfs [Simple]

476A - Dreamoon and Stairs rate : 1000 -> backtracking + some optimization [Simple]

1020B - Badge rate : 1000 -> dfs [Simple]

122A - Lucky Division rate : 1000 -> backtracking [Simple]

1324C - Frog Jumps rate : 1100 -> not so much graph but can be solved with dfs [Very Simple]

445A - DZY Loves Chessboard rate : 1200 -> can be solved with dfs [Simple(simpler than it's rating)]

217A - Ice Skating rate : 1200 -> dfs [Simple/Intermediate]

893C - Rumor rate : 1300 -> dfs [Simple]

476B - Dreamoon and WiFi rate : 1300 -> backtracking [Simple]

1638C - Inversion Graph rate : 1300 -> creativity [Simple/Intermediate]

1970C1 - Game on Tree (Easy) rate : 1300 -> trees + dfs/bfs [Simple/Intermediate]

96B - Lucky Numbers (easy) rate : 1300 -> backtracking [Simple]

1143C - Queen rate : 1400 -> trees + you can use dfs/bfs [Simple]

520B - Two Buttons rate : 1400 -> bfs [Simple/Intermediate]

580C - Kefa and Park rate : 1500 -> trees + dfs/bfs [Intermediate]

977E - Cyclic Components rate : 1500 -> dfs [Intermediate]

377A - Maze rate : 1600 -> bfs [Intermediate]

601A - The Two Routes rate : 1600 -> bfs / matrix [Intermediate/Hard]

1363C - Game On Leaves rate : 1600 -> trees / some edge cases only [Simple/Intermediate]

1144F - Graph Without Long Directed Paths rate : 1700 -> bipartite graphs [Intermediate/Hard]

1093D - Beautiful Graph rate : 1700 -> bipartite graphs [Intermediate/Hard]

1970C2 - Game on Tree (Medium) rate : 1700 -> a little hard + dfs/bfs [Intermediate/Hard]