[Codeforces][Round #736][Div.2] B : Gregor and the Pawn Game (C++)

https://codeforces.com/contest/1549/problem/B

Problem - B - Codeforces

 

[난이도] Div.2
[유형] Greedy

[풀이]
폰의 앞이 비어있다면 무조건 앞으로 전진이 가능하기 때문에
폰이 있으면 앞이 비어있는 폰은 빼주면서 정답에 이 폰의 개수를 더해줍니다.

그다음 좌측 폰부터 확인하면서
좌측 윗쪽에 폰이 있다면 이쪽으로 먼저 보내주고, 없으면 우측 윗쪽에 폰이 있는지 체크해서
폰이 있으면 이쪽으로 보내고 정답에 1을 더해줍니다.

좌측부터 검사하기 때문에 대각선 앞의 적 폰이 있는지 검사할 때 좌측->우측 순으로 검사해서 사용해 주는것이 제일 유리합니다.

#include <string>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <map>
using namespace std;
using ll = long long;

int tc,N;
string me,enemy;
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> tc;
    while(tc--){
        cin >> N >> enemy >> me;

        int ans=0;
        for(int i=0;i<N;i++){
            if(enemy[i]=='0' && me[i]=='1'){
                ans++;
                me[i]='0';
            }
        }

        for(int i=0;i<N;i++){
            if(me[i]=='0') continue;
            if(i!=0 && enemy[i-1]=='1'){
                ans++;
                continue;
            }
            if(i!=N-1 && enemy[i+1]=='1'){
                ans++;
                enemy[i+1]='0';
            }
        }
        cout << ans << '\n';
    }
}

https://github.com/has2/Problem-Solving/blob/master/codeforces/Round736-Div.2/B.cpp