Submission #1200205

#	Time	Username	Problem	Language	Result	Execution time	Memory
1200205		Ghulam_Junaid	Nafta (COI15_nafta)	C++20	100 / 100	496 ms	170344 KiB

nafta

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

typedef long long ll;

const int N = 2005;
int n, m, a[N][N];
ll dp[N][N], suff[N][N], pref[N][N];
char mat[N][N];

int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
int mx, sm;
bool vis[N][N];

bool valid(int x, int y){
    return (x > 0 and y > 0 and x <= n and y <= m and mat[x][y] != '.');
}

void dfs(int x, int y){
    vis[x][y] = 1;
    mx = max(mx, y);
    sm += mat[x][y] - '0';
    for (int i = 0; i < 4; i ++){
        int nx = x + dx[i];
        int ny = y + dy[i];

        if (!valid(nx, ny) or vis[nx][ny]) continue;
        dfs(nx, ny);
    }
}

void dnc(int j, int l, int r, int s, int e){
    int mid = (l + r) / 2, opt = s;
    if (r < l or s >= mid) return;

    dp[mid][j] = dp[s][j - 1] + pref[mid][mid] - pref[s][mid];
    for (int i = s + 1; i <= e and i < mid; i ++){
        if (dp[i][j - 1] + pref[mid][mid] - pref[i][mid] > dp[mid][j]){
            dp[mid][j] = dp[i][j - 1] + pref[mid][mid] - pref[i][mid];
            opt = i;
        }
    }

    dnc(j, l, mid - 1, s, opt);
    dnc(j, mid + 1, r, opt, e);
}

int main(){
    cin >> n >> m;
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            cin >> mat[i][j];

    for (int j = 1; j <= m; j ++){
        for (int i = 1; i <= n; i ++){
            if (vis[i][j] or mat[i][j] == '.') continue;
            mx = j, sm = 0;
            dfs(i, j);
            a[j][mx] += sm;
        }
    }

    for (int i = 1; i <= m; i ++)
        for (int j = m; j >= i; j --)
            suff[i][j] = suff[i][j + 1] + a[i][j];

    for (int i = 1; i <= m; i ++)
        for (int l = 1; l <= i; l ++)
            pref[l][i] = pref[l - 1][i] + suff[l][i];

    for (int j = 1; j <= m; j ++){
        dnc(j, 1, m, 0, m);
        for (int i = 1; i <= m; i ++)
            dp[i][j] = max(dp[i][j], dp[i - 1][j]);
    }

    for (int k = 1; k <= m; k ++)
        cout << dp[m][k] << endl;
}

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