Ligh0x74's Blog

2023-09-17发表2023-09-17更新算法 / LeetCode5 分钟读完 (大约705个字)

使数组成为递增数组的最少右移次数

直接从最小值开始判断数组是否递增，或者可以找到第一个递减的位置，然后再判断数组是否递增（因为如果数组满足条件，则其最多只有一个递减段）。

class Solution {
    public int minimumRightShifts(List<Integer> nums) {
        int n = nums.size(), idx = -1, min = 101;
        for (int i = 0; i < n; i++) {
            if (nums.get(i) < min) {
                min = nums.get(i);
                idx = i;
            }
        }
        for (int i = 0; i < n - 1; i++) {
            int x = (idx + i) % n, y = (x + 1) % n;
            if (nums.get(x) > nums.get(y)) return -1;
        }
        return (n - idx) % n;
    }
}

删除数对后的最小数组长度

贪心，比赛时我是用双指针做的，前半部分和后半部分进行匹配（当时边界想了很久，真笨！）。其他做法，参考题解：【小羊肖恩】数学 + 贪心：解决较长数组脑筋急转弯问题的关键。（因为 HashMap 很慢，所以用双指针会更快。）

方法一：贪心

class Solution {
    public int minLengthAfterRemovals(List<Integer> nums) {
        int n = nums.size(), i = 0;
        for (int j = (n + 1) / 2; j < n; j++) {
            if (nums.get(i) < nums.get(j)) {
                i++;
            }
        }
        return n - i * 2;
    }
}

方法二：贪心 + 数学

class Solution {
    public int minLengthAfterRemovals(List<Integer> nums) {
        int n = nums.size(), max = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int x : nums) {
            max = Math.max(max, map.merge(x, 1, Integer::sum));
        }
        return 2 * max <= n ? n % 2 : n - (n - max) * 2;
    }
}

统计距离为 k 的点对

枚举 \(x_{1}\oplus x_{2}\) 的值为 \(p\)，可以得到 \(y_{1}\oplus y_{2}\) 的值为 \(k-p\)。可以使用 HashMap 对前缀中的值计数来求解，需要注意循环的顺序，如果调换顺序会使代码变复杂，会花费更多的时间计算答案。

class Solution {
    public int countPairs(List<List<Integer>> coordinates, int k) {
        int ans = 0;
        Map<List<Integer>, Integer> map = new HashMap<>();
        for (var c : coordinates) {
            int x = c.get(0), y = c.get(1);
            for (int i = 0; i <= k; i++) {
                ans += map.getOrDefault(List.of(x ^ i, y ^ (k - i)), 0);
            }
            map.merge(c, 1, Integer::sum);
        }
        return ans;
    }
}

可以到达每一个节点的最少边反转次数

换根 DP，关键是要想到建立反向边，并为边添加相应的边权。

class Solution {
    public int[] minEdgeReversals(int n, int[][] edges) {
        List<int[]>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1];
            g[u].add(new int[]{v, 0});
            g[v].add(new int[]{u, 1});
        }
        int[] ans = new int[n];
        ans[0] = dfs(0, -1, g);
        dfs2(0, -1, g, ans);
        return ans;
    }
    
    private int dfs(int x, int fa, List<int[]>[] g) {
        int res = 0;
        for (int t[] : g[x]) {
            int y = t[0], w = t[1];
            if (y == fa) continue;
            res += dfs(y, x, g) + w;
        }
        return res;
    }
    
    private void dfs2(int x, int fa, List<int[]>[] g, int[] ans) {
        for (int t[] : g[x]) {
            int y = t[0], w = t[1];
            if (y == fa) continue;
            ans[y] = ans[x] + (w == 0 ? 1 : -1);
            dfs2(y, x, g, ans);
        }
    }
}

2023-09-17发表2023-09-17更新算法 / AtCoder4 分钟读完 (大约589个字)

AtCoder Beginner Contest 320

Leyland Number

模拟。

public static void solve() {
    int a = io.nextInt(), b = io.nextInt();
    int x = 1;
    for (int i = 0; i < b; i++) {
        x *= a;
    }
    int y = 1;
    for (int i = 0; i < a; i++) {
        y *= b;
    }
    io.println(x + y);
}

Longest Palindrome

模拟。

public static void solve() {
    String s = io.next();
    int n = s.length();
    int ans = 1;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            String a = s.substring(i, j + 1);
            String b = new StringBuilder(a).reverse().toString();
            if (a.equals(b)) {
                ans = Math.max(ans, j - i + 1);
            }
        }
    }
    io.println(ans);
}

Slot Strategy 2 (Easy)

暴力枚举每个转盘的时间。

public static void solve() {
    int n = 3, m = io.nextInt();
    int ans = Integer.MAX_VALUE;
    String[] s = new String[n];
    for (int i = 0; i < n; i++) {
        s[i] = io.next();
    }
    for (int i = 0; i < n * m; i++) {
        char a = s[0].charAt(i % m);
        for (int j = 0; j < n * m; j++) {
            if (i == j) continue;
            char b = s[1].charAt(j % m);
            for (int k = 0; k < n * m; k++) {
                if (i == k || j == k) continue;
                char c = s[2].charAt(k % m);
                if (a == b && b == c) {
                    ans = Math.min(ans, Math.max(i, Math.max(j, k)));
                }
            }
        }
    }
    io.println(ans == Integer.MAX_VALUE ? -1 : ans);
}

Relative Position

DFS 模拟，需要注意给的不是一棵树，所以在 DFS 时要使用访问数组，而不能用父结点。

public static void solve() {
    int n = io.nextInt(), m = io.nextInt();
    List<int[]>[] g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (int i = 0; i < m; i++) {
        int a = io.nextInt() - 1, b = io.nextInt() - 1, x = io.nextInt(), y = io.nextInt();
        g[a].add(new int[]{b, x, y});
        g[b].add(new int[]{a, -x, -y});
    }
    long[] X = new long[n];
    long[] Y = new long[n];
    Arrays.fill(X, Long.MAX_VALUE);
    Arrays.fill(Y, Long.MAX_VALUE);
    boolean[] vis = new boolean[n];
    dfs(0, vis, g, 0, 0, X, Y);
    for (int i = 0; i < n; i++) {
        if (X[i] != Long.MAX_VALUE && Y[i] != Long.MAX_VALUE) {
            io.println(X[i] + " " + Y[i]);
        } else {
            io.println("undecidable");
        }
    }
}

private static void dfs(int i, boolean[] vis, List<int[]>[] g, long x, long y, long[] X, long[] Y) {
    X[i] = x;
    Y[i] = y;
    vis[i] = true;
    for (int[] t : g[i]) {
        int j = t[0];
        if (vis[j]) continue;
        long nx = t[1] + x, ny = t[2] + y;
        dfs(j, vis, g, nx, ny, X, Y);
    }
}

Somen Nagashi

还是模拟，可以一边输入一边处理，减少代码量。

public static void solve() {
    int n = io.nextInt(), m = io.nextInt();
    long[] ans = new long[n], re = new long[n];
    PriorityQueue<Integer> q = new PriorityQueue<>();
    PriorityQueue<Integer> p = new PriorityQueue<>((a, b) -> Long.compare(re[a], re[b]));
    for (int i = 0; i < n; i++) {
        q.offer(i);
    }
    while (m-- != 0) {
        int t = io.nextInt(), w = io.nextInt(), s = io.nextInt();
        while (!p.isEmpty() && re[p.peek()] <= t) q.offer(p.poll());
        if (q.isEmpty()) continue;
        int x = q.peek();
        ans[x] += w;
        re[x] = t + s;
        p.offer(q.poll());
    }
    for (int i = 0; i < n; i++) {
        io.println(ans[i]);
    }
}

2023-09-17发表2023-09-17更新算法 / Codeforces5 分钟读完 (大约710个字)

Codeforces Round 897 (Div. 2)

green_gold_dog, array and permutation

让最小值减去最大值，就一定可以得到 \(n\) 个不同的差值。

public static void solve() {
    int n = io.nextInt();
    int[] a = new int[n];
    for (int i = 0; i < n; i++) {
        a[i] = io.nextInt();
    }
    var aux = new Integer[n];
    for (int i = 0; i < n; i++) {
        aux[i] = i;
    }
    Arrays.sort(aux, (i, j) -> a[i] - a[j]);
    int[] ans = new int[n];
    for (int i = 0; i < n; i++) {
        int t = aux[i];
        ans[t] = n - i;
    }
    for (int i = 0; i < n; i++) {
        io.print(ans[i] + " ");
    }
    io.println();
}

XOR Palindromes

算是简单的分类讨论，比赛时写的稀烂。

public static void solve() {
    int n = io.nextInt(), cnt = 0;
    char[] s = io.next().toCharArray();
    for (int i = 0; i < n / 2; i++) {
        if (s[i] != s[n - i - 1]) {
            cnt++;
        }
    }
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i <= n; i++) {
        if (i < cnt || i > n - cnt || (i - cnt) % 2 == 1 && n % 2 == 0) {
            sb.append('0');
        } else {
            sb.append('1');
        }
    }
    io.println(sb);
}

Salyg1n and the MEX Game

比赛调试一小时，疯狂超时，结果是限制太强的原因。（浪费时间。）

public static void solve() {
    int n = io.nextInt();
    int[] s = new int[n];
    for (int i = 0; i < n; i++) {
        s[i] = io.nextInt();
    }
    Arrays.sort(s);
    int x = n;
    for (int i = 0; i < n ; i++) {
        if (s[i] != i) {
            x = i;
            break;
        }
    }
    while (x != -1) {
        io.println(x);
        io.flush();
        x = io.nextInt();
    }
}

Cyclic Operations

做了一个多小时 AC，有点像之前做的内向基环树，该题每个点都有个出边，所以至少有一个环。首先要特判 \(k=1\) 的情况，该情况每个位置都必须自成一个环，即 \(a_{i}=i\)；其他情况所有环的长度都必须为 \(k\)，这样可以保证答案存在。

public static void solve() {
    int n = io.nextInt(), k = io.nextInt();
    int[] b = new int[n];
    for (int i = 0; i < n; i++) {
        b[i] = io.nextInt() - 1;
    }

    int[] in = new int[n];
    for (int i = 0; i < n; i++) {
        in[b[i]]++;
    }
    Queue<Integer> q = new LinkedList<>();
    for (int i = 0; i < n; i++) {
        if (in[i] == 0) {
            q.offer(i);
        }
    }
    while (!q.isEmpty()) {
        int x = q.poll();
        if (--in[b[x]] == 0) {
            q.offer(b[x]);
        }
    }

    int cnt = 0;
    boolean[] vis = new boolean[n];
    for (int i = 0; i < n; i++) {
        if (in[i] == 0 || vis[i]) continue;
        int len = 0;
        for ( ; !vis[i]; i = b[i]) {
            vis[i] = true;
            len++;
        }
        if (len != k) {
            io.println("NO");
            return;
        }
        cnt++;
    }
    if (k == 1 && cnt != n) {
        io.println("NO");
    } else {
        io.println("YES");
    }
}

Salyg1n and Array (simple version)

注意，\(n\) 和 \(k\) 都是偶数！手推的话可能可以做出来吧。

public static void solve() {
    int n = io.nextInt(), k = io.nextInt();
    int ans = 0, i;
    for (i = 1; i + k - 1 <= n; i += k) {
        io.println("? " + i);
        io.flush();
        ans ^= io.nextInt();
    }
    for (; i <= n; i++) {
        io.println("? " + (i - k + 1));
        io.flush();
        ans ^= io.nextInt();
    }
    io.println("! " + ans);
    io.flush();
}

Salyg1n and Array (hard version)

技巧性有点强，真想不到。就是如果多出一部分，可以通过两次异或算出来。

public static void solve() {
    int n = io.nextInt(), k = io.nextInt();
    int ans = 0, i;
    for (i = 1; i + k - 1 <= n; i += k) {
        io.println("? " + i);
        io.flush();
        ans ^= io.nextInt();
    }
    int t = n - i + 1;
    io.println("? " + (n - k + 1 - t / 2));
    io.flush();
    ans ^= io.nextInt();
    io.println("? " + (n - k + 1));
    io.flush();
    ans ^= io.nextInt();
    io.println("! " + ans);
    io.flush();
}