发表更新算法 / AtCoder9 分钟读完 (大约1297个字)

AtCoder Beginner Contest 315

tcdr

模拟。

Java

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public static void solve() {
    String s = io.next();
    var sb = new StringBuilder();
    Set<Character> set = Set.of('a', 'e', 'i', 'o', 'u');
    for (char c : s.toCharArray()) {
        if (!set.contains(c)) sb.append(c);
    }
    io.println(sb.toString());
}

C++

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void solve() {
    string s;
    cin >> s;
    s.erase(remove_if(s.begin(), s.end(), [&](char c) {
        return set{'a', 'e', 'i', 'o', 'u'}.count(c);
    }), s.end());
    cout << s << "\n";
}

The Middle Day

模拟。

Java

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public static void solve() {
    int m = io.nextInt();
    int[] d = new int[m];
    int tot = 0;
    for (int i = 0; i < m; i++) {
        d[i] = io.nextInt();
        tot += d[i];
    }
    int mid = (tot + 1) / 2;
    for (int i = 0; i < m; i++) {
        if (mid <= d[i]) {
            io.println(i + 1 + " " + mid);
            return;
        }
        mid -= d[i];
    }
}

C++

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void solve() {
    int m;
    cin >> m;
    int tot = 0;
    vector<int> d(m);
    for (int i = 0; i < m; i++) {
        cin >> d[i];
        tot += d[i];
    }
    int mid = (tot + 1) / 2;
    for (int i = 0; i < m; i++) {
        if (mid <= d[i]) {
            cout << i + 1 << " " << mid << "\n";
            return;
        }
        mid -= d[i];
    }
}

Flavors

模拟。

Java

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public static void solve() {
    int n = io.nextInt();
    List<Integer>[] buckets = new List[n + 1];
    Arrays.setAll(buckets, k -> new ArrayList<>());
    for (int i = 0; i < n; i++) {
        int f = io.nextInt(), s = io.nextInt();
        buckets[f].add(s);
    }
    int ans = 0, max1 = 0, max2 = 0;
    for (var bucket : buckets) {
        if (bucket.isEmpty()) continue;
        Collections.sort(bucket, (a, b) -> b - a);
        int a = bucket.get(0);
        if (a > max1) {
            max2 = max1;
            max1 = a;
        } else if (a > max2) {
            max2 = a;
        }
        if (bucket.size() < 2) continue;
        int b = bucket.get(1);
        ans = Math.max(ans, a + b / 2);
    }
    ans = Math.max(ans, max1 + max2);
    io.println(ans);
}

C++

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void solve() {
    int n;
    cin >> n;
    vector<vector<int>> buckets(n + 1);
    for (int i = 0; i < n; i++) {
        int f, s;
        cin >> f >> s;
        buckets[f].push_back(s);
    }
    int ans = 0, max1 = 0, max2 = 0;
    for (auto &bucket : buckets) {
        if (bucket.empty()) continue;
        nth_element(bucket.begin(), bucket.begin() + 1, bucket.end(), greater());
        if (bucket[0] > max1) {
            max2 = max1;
            max1 = bucket[0];
        } else if (bucket[0] > max2) {
            max2 = bucket[0];
        }
        if (bucket.size() < 2) continue;
        ans = max(ans, bucket[0] + bucket[1] / 2);
    }
    ans = max(ans, max1 + max2);
    cout << ans << "\n";
}

Magical Cookies

算是暴力吧。首先最多执行 \(m+n\) 次操作，然后每次操作将所有行和列遍历，判断是否可以标记。如果不优化，那么遍历的复杂度是 \(O(mn)\)，总时间复杂度就是 \(O(mn(m+n))\)，会超时。可以维护剩余的行数 \(r\) 和剩余的列数 \(c\)，那么如果某行的某颜色的数量等于列数，那么就说明可以标记该行，列同理。这样我们就可以只维护行列中的每个颜色有多少饼干，而不需要维护位置关系，从而将遍历的时间复杂度降为 \(O(26(m+n))\)。

Java

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public static void solve() {
    int m = io.nextInt(), n = io.nextInt();
    String[] arr = new String[m];
    for (int i = 0; i < m; i++) {
        arr[i] = io.next();
    }
    int[][] row = new int[m][26];
    int[][] col = new int[n][26];
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            row[i][arr[i].charAt(j) - 'a']++;
            col[j][arr[i].charAt(j) - 'a']++;
        }
    }
    int r = m, c = n;
    boolean[] vr = new boolean[m];
    boolean[] vc = new boolean[n];
    for (int k = 0; k < m + n; k++) {
        List<int[]> mr = new ArrayList<>();
        List<int[]> mc = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            if (vr[i]) continue;
            for (int j = 0; j < 26; j++) {
                if (row[i][j] == c && c >= 2) {
                    mr.add(new int[]{i, j});
                }
            }
        }
        for (int i = 0; i < n; i++) {
            if (vc[i]) continue;
            for (int j = 0; j < 26; j++) {
                if (col[i][j] == r && r >= 2) {
                    mc.add(new int[]{i, j});
                }
            }
        }
        for (int[] p : mr) {
            r--;
            vr[p[0]] = true;
            for (int i = 0; i < n; i++) {
                col[i][p[1]]--;
            }
        }
        for (int[] p : mc) {
            c--;
            vc[p[0]] = true;
            for (int i = 0; i < m; i++) {
                row[i][p[1]]--;
            }
        }
    }
    io.println(r * c);
}

Prerequisites

首先找到第 \(1\) 本书的所有前置书，然后对所有书进行拓扑排序，将之前找到的前置书按拓扑排序的倒序打印即可。或者直接 DFS。。

Java

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public static void solve() {
    int n = io.nextInt();
    int[] indegree = new int[n];
    List<Integer>[] g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (int i = 0; i < n; i++) {
        int c = io.nextInt();
        for (int j = 0; j < c; j++) {
            int q = io.nextInt() - 1;
            g[i].add(q);
            indegree[q]++;
        }
    }

    boolean[] mark = new boolean[n];
    Queue<Integer> q = new LinkedList<>();
    q.offer(0);
    while (!q.isEmpty()) {
        int x = q.poll();
        for (int y : g[x]) {
            if (mark[y]) continue;
            mark[y] = true;
            q.offer(y);
        }
    }

    for (int i = 0; i < n; i++) {
        if (indegree[i] == 0) {
            q.offer(i);
        }
    }
    Deque<Integer> ans = new ArrayDeque<>();
    while (!q.isEmpty()) {
        int x = q.poll();
        if (mark[x]) ans.push(x);
        for (int y : g[x]) {
            if (--indegree[y] == 0) {
                q.offer(y);
            }
        }
    }
    while (!ans.isEmpty()) io.print(ans.pop() + 1 + " ");
    io.println();
}

C++

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void solve() {
    int n;
    cin >> n;

    vector<vector<int>> adj(n);
    for (int i = 0; i < n; i++) {
        int c;
        cin >> c;
        for (int j = 0; j < c; j++) {
            int q;
            cin >> q;
            q--;
            adj[i].push_back(q);
        }
    }

    vector<bool> mark(n);
    auto dfs = [&](auto self, int x) {
        if (mark[x]) {
            return;
        }
        for (auto y : adj[x]) {
            self(self, y);
        }
        mark[x] = true;
        if (x != 0) {
            std::cout << x + 1 << " ";
        }
    };
    dfs(dfs, 0);
    cout << "\n";
}

Shortcuts

动态规划，调试好久。。如果所有点都选，那么答案最多为 \(10^{9}\)，所以可以确定不选的点不会超过 \(30\)。然后定义状态 \(dp[i][j]\) 表示到达第 \(i\) 个点并且总共跳过 \(j\) 个点的最短距离。如何想到定义该状态呢，因为答案和具体选哪几个点无关，只和最短距离以及跳过多少个点有关，大概是这样吧。

Java

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public static void solve() {
    int c = 30;
    int n = io.nextInt();
    int[] x = new int[n];
    int[] y = new int[n];
    for (int i = 0; i < n; i++) {
        x[i] = io.nextInt();
        y[i] = io.nextInt();
    }
    double[][] dp = new double[n][c];
    for (int i = 0; i < n; i++) {
        Arrays.fill(dp[i], Integer.MAX_VALUE);
    }
    dp[0][0] = 0;
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < c; j++) {
            for (int k = i + 1; k < n && k - i - 1 + j < c; k++) {
                int nj = j + k - i - 1;
                dp[k][nj] = Math.min(dp[k][nj], dp[i][j] + Math.sqrt((x[i] - x[k]) * (x[i] - x[k]) + (y[i] - y[k]) * (y[i] - y[k])));
            }
        }
    }
    double ans = Integer.MAX_VALUE;
    for (int i = 0; i < c; i++) {
        ans = Math.min(ans, dp[n - 1][i] + (i == 0 ? 0 : 1 << (i - 1)));
    }
    io.println(ans);
}

AtCoder Beginner Contest 315

https://ligh0x74.github.io/2023/08/21/AtCoder Beginner Contest 315/

作者

Ligh0x74

发布于

2023-08-21

更新于

2023-08-21

许可协议

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