1680B - Robots - CodeForces Solution

implementation *800

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Python Code: 

t = int(input())
for i in range(t):
    n, m = map(int, input().split())
    s = []
    for j in range(n):
        s.append(input())
    minx = 10 ** 9
    miny = 10 ** 9
    for x in range(n):
        for y in range(m):
            if s[x][y] == 'R':
                minx = min(minx, x)
                miny = min(miny, y)
    print('YES' if s[minx][miny] == 'R' else 'NO')

C++ Code: 

#include <bits/stdc++.h>     
using namespace std;
#define ll long long
 
ll MOD=1000000007;
 
// vector<ll> seive(ll n){ //vector of primes in [1,n]
//     vector <ll> isPrime(n+1,1);
//     vector <ll> temp;
//     isPrime[0]=isPrime[1]=0;
//     for(ll i=2;i<n+1;i++){
//         if(isPrime[i]){
//             for(ll j=i*i;j<n+1;j+=i){
//                 isPrime[j]=0;
//             }
//         }
//     }
//     for(ll i=0;i<n+1;i++){
//         if(isPrime[i]) temp.push_back(i);
//     }
//     return temp;
// }
 
// ll power(ll a,ll b){
//     ll ans=1;
//     while(b){
//         if(b&1) ans*=a;
//         a=a*a;
//         b=b>>1;
//     }
//     return ans;
// }
void solve(){
    ll n,m;
    cin >> n >> m;
    char a[n][m];
    ll mx=INT_MAX,my=INT_MAX;
    ll rf=0;
    for(ll i=0;i<n;i++){
        for(ll j=0;j<m;j++){
            cin >> a[i][j];
            if(a[i][j]=='R'){
                rf=1;
                mx=min(mx,i);
                my=min(my,j);
            }
        }
    }

    if(rf==0) {cout << "NO" << endl; return;}
    if(a[mx][my]=='R') cout << "YES" << endl;
    else cout << "NO" << endl;
}
 
int main(){  
    ios_base::sync_with_stdio(false);                         
    cin.tie(NULL); 
    // solve(); 
    ll t; cin >> t;
    while(t--){
        solve();
    }
}


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