AtCoder Beginner Contest 188 题解

AtCoder Beginner Contest 188

A，B很简单就不多说

C - ABC Tournament

找出前一半的最大值和后一半的最大值，二者中较小的那一个对应的序号就是最后的答案。

时间复杂度：$\mathcal{O}(2^N)$

using ll = long long;
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    ll a[70000];
    int n;
    cin >> n, n = pow(2, n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    ll m1 = a[0], idx = 0;
    for (int i = 1; i < n / 2; ++i)
        if (a[i] > m1) m1 = a[i], idx = i;
    ll m2 = a[n / 2], idxx = n / 2;
    for (int i = n / 2 + 1; i < n; ++i)
        if (a[i] > m2) m2 = a[i], idxx = i;
    cout << (m1 < m2 ? idx + 1 : idxx + 1) << "\n";
    return 0;
}

D - Snuke Prime

时间复杂度：$\mathcal{O}(N)$

using ll = long long;
int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    ll N, C;
    cin >> N >> C;
    vector<pair<ll, ll>> event;
    for (ll i = 0; i < N; ++i) {
        ll a, b, c;
        cin >> a >> b >> c;
        event.push_back({a - 1, c});
        event.push_back({b, -c});
    }
    sort(event.begin(), event.end());
    ll ans = 0, fee = 0, time = 0;
    for (auto [x, y] : event) {
        if (x != time) {
            ans += min(C, fee) * (x - time);
            time = x;
        }
        fee += y;
    }
    cout << ans << "\n";
    return 0;
}

E - Peddler

题意：N 个城镇，每个镇子购买和卖出 $1kg$ 黄金的价格是 $A_i$ ，现在可以我们有 Ｍ条道路，问怎么才能获得最大收益

思路：虽然看过去需要搜索做，但这里只要逆向动态规划即可。

int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int N, M;
    cin >> N >> M;
    vector<int> A(N + 1);
    for (int i = 1; i <= N; ++i) cin >> A[i];

    vector<vector<int>> e(N + 1);
    for (int i = 0; i < M; ++i) {
        int u, v;
        cin >> u >> v;
        // e[u].emplace_back(v), e[v].emplace_back(u);
        e[u].emplace_back(v);  // 这里单向和双向都没什么区别
    }

    vector<int> h(N + 1, -1e9);
    int ans = -1e9;
    for (int i = N; i > 0; --i) {
        for (auto v : e[i]) h[i] = max(h[i], h[v]);
        ans = max(ans, h[i] - A[i]);
        h[i] = max(h[i], A[i]);
    }
    cout << ans << "\n";
    return 0;
}

F - +1-1x2

• 如果 $X\ge Y$ ，则答案为$X-Y$
• 如果 $X<Y$ ，则可以考虑用ＤＦＳ，详细可以看代码

using ll = long long;
ll x, y, ans = 1e18;
map<ll, ll> mp;
ll dfs(ll a) {
    if (mp.count(a)) return mp[a];
    if (a <= x) return x - a;
    if (a & 1)
        return mp[a] = min(min(dfs((a + 1) / 2), dfs((a - 1) / 2)) + 2, a - x);
    else
        return mp[a] = min(dfs(a / 2) + 1, a - x);
}
int main() {
    cin >> x >> y;
    cout << dfs(y);
}