Submission #947992

#	Time	Username	Problem	Language	Result	Execution time	Memory
947992		BhavayGoyal	Nafta (COI15_nafta)	C++14	34 / 100	1092 ms	49088 KiB

nafta

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

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

template<class T> using oset = 
            tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define ll long long
#define ld long double
#define ar array
#define vi vector<int>
#define vii vector<vector<int>>
#define pii pair<int, int>
#define pb push_back
#define all(x) x.begin(), x.end()
#define f first
#define s second
#define endl "\n"

const int MOD = 1e9+7;
const int inf = 1e9;
const ll linf = 1e18;

const int d4i[4]={-1, 0, 1, 0}, d4j[4]={0, 1, 0, -1};
const int d8i[8]={-1, -1, 0, 1, 1, 1, 0, -1}, d8j[8]={0, 1, 1, 1, 0, -1, -1, -1};

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


// -------------------------------------------------- Main Code --------------------------------------------------

const int N = 2005;

char arr[N][N];
int n, m, Cost[N][N], dp[N][N];
set<int> temp;

int dfs(int i, int j) {
    temp.insert(j);
    int cost = arr[i][j]-'0';
    arr[i][j] = '.';
    for (int x = 0; x < 4; x++) {
        int ni = d4i[x]+i, nj = d4j[x]+j;
        if (ni >= 1 && ni <= n && nj >= 1 && nj <= m && arr[ni][nj] != '.') {
            cost += dfs(ni, nj);
        }
    }
    return cost;
}

void sol() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> arr[i][j];
        }
    }

    vii cost1(m+1, vi(m+1));
    for (int j = 1; j <= m; j++) {
        for (int i = 1; i <= n; i++) {
            if (arr[i][j] == '.') continue;
            temp.clear();
            int cost = dfs(i, j);
            for (auto &x : temp) {
                // starting from j, x.s is a part of it
                cost1[j][x] += cost;
            }
        }
    }

    for (int i = 1; i <= m; i++) for (int j = 1; j <= m; j++) cost1[i][j] += cost1[i-1][j];
    for (int i = 0; i <= m; i++) for (int j = 0; j <= m; j++) Cost[i][j] = cost1[j][j] - cost1[i][j];

    for (int k = 1; k <= m; k++) {
        for (int i = 0; i <= m; i++) {
            for (int j = i; j >= 0; j--) {
                // current drill on i, last drill on j
                dp[i][k] = max(dp[i][k], dp[j][k-1] + Cost[j][i]);
            }
        }

        int ans = 0; 
        for (int i = 0; i <= m; i++) ans = max(ans, dp[i][k]);
        cout << ans << endl;
    }
}

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

    int t = 1;
    // cin >> t; 
    while (t--) {
        sol();
    }
    return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 23 / 23
• Subtask 3: 0 / 66