Submission #227917

#	Time	Username	Problem	Language	Result	Execution time	Memory
227917		kartel	Strah (COCI18_strah)	C++14	110 / 110	269 ms	24000 KiB

strah

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define in(x) freopen(x, "r", stdin)
#define out(x) freopen(x, "w", stdout)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#define F first
#define S second
#define pb push_back
#define N +100500
#define M ll(1e9 + 7)
#define sz(x) (int)x.size()
#define re return
#define oo ll(1e18)
#define el '\n'
using namespace std;
//using namespace __gnu_pbds;
//typedef tree <int, null_type, less_equal <int> , rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef long long ll;
typedef long double ld;

ll ans;
int h[2005][2005], le[2005], ri[2005], i, j, n, m;
vector <int> st;
char c[2005][2005];

int main()
{
    srand(time(0));
    ios_base::sync_with_stdio(0);
    iostream::sync_with_stdio(0);
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);

//    in("sum.in");
//    out("output.txt");

    cin >> n >> m;
    for (i = 1; i <= n; i++)
      for (j = 1; j <= m; j++) cin >> c[i][j];

    for (j = 1; j <= m; j++)
    {
        for (i = 1; i <= n; i++)
            if (c[i][j] == '#') h[i][j] = 0;
                         else   h[i][j] = h[i - 1][j] + 1;
    }

    for (i = 1; i <= n; i++)
    {
        st.clear();

        for (j = 1; j <= m; j++)
        {
            while (st.size() && h[i][st.back()] >= h[i][j]) st.pop_back();

            if (st.size()) le[j] = st.back();
                      else le[j] = 0;

            st.pb(j);
        }

        st.clear();

        for (j = m; j >= 1; j--)
        {
            while (st.size() && h[i][st.back()] > h[i][j]) st.pop_back();

            if (st.size()) ri[j] = st.back();
                      else ri[j] = m + 1;

            st.pb(j);
        }

        for (j = 1; j <= m; j++)
        {
            ll k1 = j - le[j];
            ll k2 = ri[j] - j;

            ll cur = k1 * (k2 * (k2 + 1) / 2) + k2 * (k1 * (k1 + 1) / 2) - k1 * k2;

            ans += cur * h[i][j] * (h[i][j] + 1) / 2;
//            cout << cur << el;
        }
//        cout << el;
    }
    cout << ans;
}



// 119 78 175 411

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 11 / 11
• Subtask 5: 11 / 11
• Subtask 6: 11 / 11
• Subtask 7: 11 / 11
• Subtask 8: 11 / 11
• Subtask 9: 11 / 11
• Subtask 10: 11 / 11