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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#194510#7523. Partially Free Mealucup-team228#TL 2ms13900kbC++203.3kb2023-09-30 20:57:372023-09-30 20:57:39

Judging History

你现在查看的是最新测评结果

  • [2023-09-30 20:57:39]
  • 评测
  • 测评结果:TL
  • 用时:2ms
  • 内存:13900kb
  • [2023-09-30 20:57:37]
  • 提交

answer

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("avx2") // doesn't work on Yandex
//#pragma GCC target("avx") // for old judges
//#pragma GCC target("bmi,bmi2,lzcnt,popcnt") // fast bit operations

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

template <typename T>
using min_heap = priority_queue<T, vector<T>, greater<>>;

const ll inf = 1e18;
const int N = 2e5 + 10;
ll a[N], b[N];

ll dp[2][N];

pair<ll, ll> c[N];
ll ans[N];

void solve(int n) {
    for (int i = 1; i <= n; i++) {
        c[i] = {b[i], a[i]};
    }
    sort(c + 1, c + n + 1);
    for (int i = 1; i <= n; i++) {
        a[i] = c[i].second;
        b[i] = c[i].first;
    }
    for (int i = 1; i <= n; i++) {
        dp[0][i] = b[i];
    }
    for (int k = 1; k <= n; k++) {
        dp[k & 1][k - 1] = inf;
        ans[k] = inf;
        for (int i = k; i <= n; i++) {
            dp[k & 1][i] = min(dp[k & 1][i - 1] - b[i - 1] + b[i], dp[(k & 1) ^ 1][i - 1] - b[i - 1] + a[i] + b[i]);
            ans[k] = min(ans[k], dp[k & 1][i]);
        }
    }
}

ll res[N], mem_b[N], mem_mask[N];

void slow(int n) {
    for (int i = 1; i <= n; i++) {
        res[i] = inf;
    }
    for (int mask = 1; mask < (1 << n); mask++) {
        ll sum = 0, mx = 0;
        for (int i = 1; i <= n; i++) {
            if (mask & (1 << (i - 1))) {
                sum += a[i];
                mx = max(mx, b[i]);
            }
        }
        int k = __builtin_popcount(mask);
        if (res[k] > sum + mx) {
            res[k] = sum + mx;
            mem_b[k] = mx;
            mem_mask[k] = mask;
        } else if (res[k] == sum + mx && mem_b[k] > mx) {
            mem_b[k] = mx;
            mem_mask[k] = mask;
        }
    }
}

void stress() {
    mt19937 rnd;
    while (true) {
        int n = rnd() % 3 + 1;
        for (int i = 1; i <= n; i++) {
            a[i] = rnd() % 10 + 1;
            b[i] = rnd() % 10 + 1;
        }
        solve(n);
        slow(n);
        bool ok = true;
//        for (int i = 1; i + 1 <= n; i++) {
//            ok &= mem_b[i] <= mem_b[i + 1];
//        }
        for (int i = 1; i <= n; i++) {
            ok &= ans[i] == res[i];
        }
        if (ok) {
            cout << "OK" << endl;
        } else {
            cout << "WA\n";
            cout << n << "\n";
            for (int i = 1; i <= n; i++) {
                cout << a[i] << " " << b[i] << "\n";
            }
            cout << "\n";
            for (int i = 1; i <= n; i++) {
                cout << ans[i] << "\n";
            }
            cout << "\n";
            for (int i = 1; i <= n; i++) {
                cout << res[i] << "\n";
            }
            break;
        }
    }
    exit(0);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
#endif

    // stress();

    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i] >> b[i];
    }
    solve(n);
    for (int i = 1; i <= n; i++) {
        cout << ans[i] << "\n";
    }

//    cout << "\n";
//    slow(n);
//    for (int i = 1; i <= n; i++) {
//        cout << res[i] << "\n";
//    }

#ifdef LOCAL
    cout << "\nTime elapsed: " << double(clock()) / CLOCKS_PER_SEC << " s.\n";
#endif
}

詳細信息

Test #1:

score: 100
Accepted
time: 2ms
memory: 13900kb

input:

3
2 5
4 3
3 7

output:

7
11
16

result:

ok 3 lines

Test #2:

score: -100
Time Limit Exceeded

input:

200000
466436993 804989151
660995237 756645598
432103296 703610564
6889895 53276988
873617076 822481192
532911431 126844295
623111499 456772252
937464699 762157133
708503076 786039753
78556972 5436013
582960979 398984169
786333369 325119902
930705057 615928139
924915828 506145001
164984329 208212435...

output:


result: