제출 #1250907

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1250907		NoLove	Nafta (COI15_nafta)	C++20	0 / 100	14 ms	32328 KiB

nafta

/**
 *    author : Lăng Trọng Đạt
 *    created: 02-08-2025 
**/
#include <bits/stdc++.h>
using namespace std;
#ifndef LANG_DAT
#define db(...) ;
#endif // LANG_DAT
#define int long long
#define f first
#define se second
#define pb push_back
#define all(v) (v).begin(), (v).end()
#define FOR(i, a, b) for (int i = (a); (i) <= (b); (i++))
#define FD(i, lo, up) for (int i = (up); (i) >= (lo); i--)
#define si(x) (int)(x.size())
bool mx(int& a, int b) { if (b > a) {a = b; return true;} return false;}
bool mi(int& a, int b) { if (b < a) {a = b; return true;} return false;}
using pii = pair<int, int>;
using vi = vector<int>;
const int INF = 1e18 + 5;
const int MOD = 1e9 + 7;

const int N = 2e3 + 5;
int g[N][N];
int n, m, k, q, a, b, c;

int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int comp[N][N], sum[N * N], trai[N * N], comp_cnt = 0;
vector<pii> oil[N];
void pre() {
    FOR(i, 1, n) FOR(j, 1, m) {
        if (comp[i][j] == 0 && g[i][j] != -1) {
            trai[++comp_cnt] = m + 1;
            queue<pii> q; 
            q.push({i, j});
            while (si(q)) {
                auto[x, y] = q.front(); q.pop();
                if (1 <= x && x <= n && 1 <= y && y <= m && g[x][y] != -1 && !comp[x][y]) {
                    comp[x][y] = comp_cnt;
                    sum[comp_cnt] += g[x][y]; 
                    mi(trai[comp_cnt], y);

                    FOR(k, 0, 3) {
                        q.push({x + dx[k], y + dy[k]});
                    }   
                }
            }
        }
    }
    FOR(i, 1, m) {
        vector<vector<int>> cur;
        FOR(x, 1, n) {
            if (g[x][i] != -1) {
                cur.push_back({trai[comp[x][i]], comp[x][i], sum[comp[x][i]]});
            }
        }
        sort(all(cur));
        cur.resize(unique(all(cur)) - cur.begin());

        int add = 0;
        for(auto v : cur) {
            oil[i].push_back({v[0], v[2]});
        }
        for (int j = si(oil[i]) - 2; j >= 0; j--) {
            oil[i][j].se += oil[i][j + 1].se;
        }
    }
}

int dp[N][N]; // dp[i][j]: nếu khoan j cái và cái cuối cùng ở vị trí <= i thì tối đa là ?
int opt[N]; // opt[l] <= opt[mid] <= opt[r]
void solve(int l, int r, int khoan) {
    if (r - l > 1) {
        int mid = (l + r) / 2, add = 0;
        auto id = prev(upper_bound(all(oil[mid]), make_pair(opt[r], INF))) - oil[mid].begin();
        // db(l, r, mid, oil[mid][id], opt[l], opt[r])
        for (int i = id; i >= 0 && oil[mid][i].first >= opt[l]; i--) {
            if (mx(dp[khoan][mid], dp[khoan - 1][oil[mid][i].f - 1] + oil[mid][i].se)) {
                opt[mid] = oil[mid][i].f;
            }
            // if (mid == 5 && khoan == 2) {
                // db(mid, dp[khoan - 1][oil[mid][i].f - 1], oil[mid][i]);
            // }
        }

        solve(l, mid, khoan);
        solve(mid, r, khoan);
    }
}
void trau() {
    FOR(j, 1, m) {
        FOR(i, 1, m) {
            dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            vector<vector<int>> oil;
            FOR(x, 1, n) {
                if (g[x][i] != -1) {
                    oil.push_back({trai[comp[x][i]], comp[x][i], sum[comp[x][i]]});
                }
            }
            sort(all(oil));
            oil.resize(unique(all(oil)) - oil.begin());
            reverse(all(oil));

            int add = 0;
            // for (auto[left_most, id, tong] : oil) {
            for (auto& cur : oil) {
                add += cur[2];
                mx(dp[i][j], dp[cur[0] - 1][j - 1] + add);
            }
            db(i, j, dp[i][j])
        }

        cout << dp[m][j] << "\n";
    }
}
int32_t main() {
    cin.tie(0)->sync_with_stdio(0);
    if (fopen("hi.inp", "r")) {
        freopen("hi.inp", "r", stdin);
//        freopen("hi.out", "w", stdout);
    } 

    cin >> n >> m;
    FOR(i, 1, n) FOR(j, 1, m) {
        char c; cin >> c;
        g[i][j] = (c == '.' ? -1 : c - '0');
    }

    memset(dp, -1, sizeof(dp));
    FOR(i, 0, m) dp[0][i] = 0;
    pre();
    opt[0] = 0, opt[m + 1] = m + 1;
    FOR(i, 1, m) {
        dp[i - 1][0] = 0;
        FOR(j, 1, m) opt[j] = m + 1;
        // db(i, oil[i])
        solve(0, m + 1, i);
        FOR(j, 1, m) {
            mx(dp[i][j], dp[i][j - 1]);
            // db(i, j, dp[i][j])
        }
        cout << dp[i][m] << "\n";
    }
}

컴파일 시 표준 에러 (stderr) 메시지

nafta.cpp: In function 'int32_t main()':
nafta.cpp:122:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  122 |         freopen("hi.inp", "r", stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~

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