1452A - Robot Program.cpp

/*
There is an infinite 2-dimensional grid. The robot stands in cell (0,0) and wants to reach cell (x,y). Here is a list of possible commands the robot can execute:

move north from cell (i,j) to (i,j+1);
move east from cell (i,j) to (i+1,j);
move south from cell (i,j) to (i,j−1);
move west from cell (i,j) to (i−1,j);
stay in cell (i,j).
The robot wants to reach cell (x,y) in as few commands as possible. However, he can't execute the same command two or more times in a row.

What is the minimum number of commands required to reach (x,y) from (0,0)?

Input
The first line contains a single integer t (1≤t≤100) — the number of testcases.

Each of the next t lines contains two integers x and y (0≤x,y≤104) — the destination coordinates of the robot.

Output
For each testcase print a single integer — the minimum number of commands required for the robot to reach (x,y) from (0,0) if no command is allowed to be executed two or more times in a row.

Example
inputCopy
5
5 5
3 4
7 1
0 0
2 0
outputCopy
10
7
13
0
3
Note
The explanations for the example test:

We use characters N, E, S, W and 0 to denote going north, going east, going south, going west and staying in the current cell, respectively.

In the first test case, the robot can use the following sequence: NENENENENE.

In the second test case, the robot can use the following sequence: NENENEN.

In the third test case, the robot can use the following sequence: ESENENE0ENESE.

In the fourth test case, the robot doesn't need to go anywhere at all.

In the fifth test case, the robot can use the following sequence: E0E.


*/












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

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin>>t;
    while(t--){
        int x, y;
        cin>>x>>y;

        if(x==y) cout<<x+y<<"\n";
        else {
            int a=0, b=0;
            char dir, prev;
            if(x>y){
                dir='E';
                prev='O';
            } else {
                dir='N';
                prev='O';
            }
            int step=0;

            while(a<x || b<y){
                if(dir=='E' && a<x && (prev=='N' || prev=='O')){
                    a++;
                    dir='N';
                    prev='E';
                    step++;
                }
                else if(dir=='N' && b<y && (prev=='E' || prev=='O')){
                    b++;
                    dir='E';
                    prev='N';
                    step++;
                } else if(b==y && dir=='N' ){
                    step++;
                    prev='O';
                    dir='E';
                } else if(a==x && dir=='E'){
                    step++;
                    dir='N';
                    prev='O';
                }

            }

            cout<<step<<"\n";
        }


    }


    return 0;
}