Problem: Example

Input

2 1
WE


Output

1 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define M_PI 3.14159265358979323846

using namespace std;

//typedef
//-------------------------#include <bits/stdc++.h>

#define M_PI 3.14159265358979323846

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s)
{
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}
template <class T>
inline string toString(T x)
{
    ostringstream sout;
    sout << x;
    return sout.str();
}
inline int readInt()
{
    int x;
    scanf("%d", &x);
    return x;
}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;

//container util

//------------------------------------------
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a) * (a))
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define SORT(c) sort((c).begin(), (c).end())

//repetition
//------------------------------------------
#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define REP(i, n) FOR(i, 0, n)
//#define MOD 1000000007

#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define trav(a, x) for (auto &a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const double EPS = 1E-8;

#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
const int INF = 2147483647;

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit()
{
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);

    int n, m;
    cin >> n >> m;

    vector<string> s;
    for (int i = 0; i < m; i++)
    {
        string t;
        cin >> t;
        s.push_back(t);
    }

    int cnt1 = 0, cnt2 = 0;
    for (int j = 0; j < n; j++)
    {
        for (int i = 0; i < m; i++)
        {
            if (s[i][j] == 'W')
            {
                cnt2++;
            }
        }
    }
    int lb = 0, ub = 1;
    int mi = cnt1 + cnt2;

    for (int j = 0; j < n; j++)
    {

        for (int i = 0; i < m; i++)
        {
            if (s[i][j] == 'W')
            {
                cnt2--;
            }
            else if (s[i][j] == 'E')
            {
                cnt1++;
            }
        }
        if (mi > cnt1 + cnt2)
        {
            mi = cnt1 + cnt2;
            lb = j + 1, ub = j + 2;
        }
    }

    int cnt = 0;
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (s[i][j] == 'E')
            {
                cnt++;
            }
        }
    }
    if (mi > cnt)
    {
        lb = n, ub = n + 1;
    }
    cout << lb << " " << ub << endl;
    return 0;
}