Ligh0x74's Blog

2023-10-02发表2023-10-02更新算法 / LeetCode6 分钟读完 (大约863个字)

有序三元组中的最大值 II

方法一：枚举 j

比赛时第一想法是枚举 \(j\)，然后取左边和右边的最大值计算答案。

class Solution {
    public long maximumTripletValue(int[] nums) {
        int n = nums.length;
        int[] pre = new int[n + 1];
        int[] suf = new int[n + 1];
        for (int i = 0; i < n; i++) {
            pre[i + 1] = Math.max(pre[i], nums[i]);
        }
        for (int i = n - 1; i >= 0; i--) {
            suf[i] = Math.max(suf[i + 1], nums[i]);
        }
        long ans = 0L;
        for (int j = 1; j < n - 1; j++) {
            ans = Math.max(ans, (long) (pre[j] - nums[j]) * suf[j + 1]);
        }
        return ans;
    }
}

方法二：枚举 k

参考灵神的题解，可以枚举 \(k\)，使空间复杂度降为 \(O(1)\)。主要想法就是枚举时，维护前缀最大差值。

class Solution {
    public long maximumTripletValue(int[] nums) {
        long ans = 0;
        int maxDiff = 0, preMax = 0;
        for (int x : nums) {
            ans = Math.max(ans, (long) maxDiff * x);
            maxDiff = Math.max(maxDiff, preMax - x);
            preMax = Math.max(preMax, x);
        }
        return ans;
    }
}

无限数组的最短子数组

比赛时思路很快出来，但是实现的时候漏掉一些边界条件，导致 WA 多次，本来有机会进第一页的。比较直接的想法是枚举起始位置，然后利用前缀和二分结束的位置，如果满足条件就记入答案。也可以使用哈希表来优化，避免二分。这里贴一下灵神的滑动窗口解法，时间复杂度 \(O(n)\)，空间复杂度 \(O(1)\)。

class Solution {
    public int minSizeSubarray(int[] nums, int target) {
        int n = nums.length;
        long sum = 0L, pre = 0L;
        for (int x : nums) sum += x;
        int q = (int) (target / sum), r = (int) (target % sum);
        int lo = 0, hi = 0, ans = Integer.MAX_VALUE;
        while (hi < 2 * n) {
            pre += nums[hi++ % n];
            while (pre > r) {
                pre -= nums[lo++ % n];
            }
            if (pre == r) {
                ans = Math.min(ans, hi - lo);
            }
        }
        return ans == Integer.MAX_VALUE ? -1 : ans + q * n;
    }
}

有向图访问计数

看完题目就知道是内向基环树，直接修改上次的代码，轻松通过，本题是上次的简化版。

class Solution {
    public int[] countVisitedNodes(List<Integer> edges) {
        int n = edges.size();
        // 建立环外节点的反向边
        int[] in = new int[n];
        List<Integer>[] reverse = new List[n];
        Arrays.setAll(reverse, r -> new ArrayList<>());
        for (int i = 0; i < n; i++) {
            in[edges.get(i)]++;
            reverse[edges.get(i)].add(i);
        }
        // 拓扑序去除环外节点
        Queue<Integer> q = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (in[i] == 0) q.add(i);
        }
        while (!q.isEmpty()) {
            int x = q.poll();
            if (--in[edges.get(x)] == 0) {
                q.offer(edges.get(x));
            }
        }
        // 记录每个节点在哪个环
        int[] cirNum = new int[n];
        boolean[] vis = new boolean[n];
        List<Integer> circles = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (!vis[i] && in[i] != 0) {
                int cnt = 0;
                for (int cur = i; !vis[cur]; cur = edges.get(cur)) {
                    vis[cur] = true;
                    cirNum[cur] = circles.size();
                    cnt++;
                }
                circles.add(cnt);
            }
        }
        // 对环内的每个节点向环外进行 dfs
        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            if (in[i] != 0) dfs(i, reverse, in, circles.get(cirNum[i]), ans);
        }
        return ans;
    }
    
    private void dfs(int x, List<Integer>[] reverse, int[] in, int len, int[] ans) {
        ans[x] = len;
        for (int y : reverse[x]) {
            if (in[y] != 0) continue;
            dfs(y, reverse, in, len + 1, ans);
        }
    }
}

有个非常简单的写法，参考题解。

class Solution {
    public int[] countVisitedNodes(List<Integer> edges) {
        int n = edges.size();
        int[] ans = new int[n];
        int[] vis = new int[n];
        for (int i = 0; i < n; i++) {
            if (ans[i] == 0) {
                int cnt = 0, cirLen = 0, totLen = 0, pos = i;
                while (vis[pos] == 0) {
                    vis[pos] = ++cnt;
                    pos = edges.get(pos);
                }
                if (ans[pos] == 0) {
                    cirLen = cnt - vis[pos] + 1;
                    totLen = cnt;
                } else {
                    totLen = cnt + ans[pos];
                }
                pos = i;
                while (ans[pos] == 0) {
                    ans[pos] = Math.max(totLen--, cirLen);
                    pos = edges.get(pos);
                }
            }
        }
        return ans;
    }
}

2023-10-02发表2023-10-02更新算法 / Codeforces5 分钟读完 (大约718个字)

Codeforces Round 901 (Div. 2)

Jellyfish and Undertale

每个工具对答案的贡献为 \(\min(a-1,x_{i})\)，并且答案可能会爆 \(long\)，状态好差，WA 两次。

public static void solve() {
    int a = io.nextInt(), b = io.nextInt(), n = io.nextInt();
    long ans = b;
    for (int i = 0; i < n; i++) {
        ans += Math.min(a - 1, io.nextInt());
    }
    io.println(ans);
}

Jellyfish and Game

数学题。当 \(k\bmod 2=1\) 时，先手会得到 \(\min(0,maxb-mina)\)；反之，后手此时必然有最小价值的苹果 \(\min(mina,minb)\)，而先手此时会有最大价值的苹果 \(\max(maxa,maxb)\)，做一次交换即可。

public static void solve() {
    int n = io.nextInt(), m = io.nextInt(), k = io.nextInt();

    long suma = 0L;
    int mina = Integer.MAX_VALUE, maxa = Integer.MIN_VALUE;
    for (int i = 0; i < n; i++) {
        int x = io.nextInt();
        suma += x;
        mina = Math.min(mina, x);
        maxa = Math.max(maxa, x);
    }

    int minb = Integer.MAX_VALUE, maxb = Integer.MIN_VALUE;
    for (int i = 0; i < m; i++) {
        int x = io.nextInt();
        minb = Math.min(minb, x);
        maxb = Math.max(maxb, x);
    }

    if (k % 2 == 1) {
        io.println(suma + Math.max(0, maxb - mina));
    } else {
        io.println(suma + Math.max(0, maxb - mina) - Math.max(maxa, maxb) + Math.min(mina, minb));
    }
}

Jellyfish and Green Apple

完了，官方解法看不懂，下面是我的解法，就是一直乘二求余，贪心的拆分。

public static void solve() {
    int n = io.nextInt(), m = io.nextInt();

    int x = n % m, y = m / gcd(n, m);
    if ((y & (y - 1)) != 0) {
        io.println(-1);
        return;
    }

    long ans = 0L;
    while (x != 0) {
        ans += x;
        x *= 2;
        x %= m;
    }
    io.println(ans);
}

private static int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

下面是官解，大概懂了。\(\frac{a}{b}\) 表示最终每个人得到的苹果价值的小数部分（可以表示为 \(\sum_{i\in S}\frac{1}{2^{i}}\)），因为 \(b\) 必须是二的幂，所以 \(a\) 中 \(1\) 的个数就表示这个人所需的最小苹果片数（就是集合 \(S\) 中元素的个数），乘以 \(m\) 就表示最后总共的苹果片数，然后减去最开始的 \(n\) 片就是需要进行操作的次数（因为每次操作会增加 \(1\) 片）。（这个题解很详细，提供了另一种视角）

public static void solve() {
    int n = io.nextInt(), m = io.nextInt();

    n %= m;
    int x = gcd(n, m), a = n / x, b = m / x;
    if ((b & (b - 1)) != 0) {
        io.println(-1);
    } else {
        io.println((long) Integer.bitCount(a) * m - n);
    }
}

private static int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

Jellyfish and Mex

其实是很简单的动态规划，但就是动不起来啊。定义 \(dp[i]\) 为将 \(MEX\) 变为 \(i\) 需要的最小代价，然后从大到小转移即可。

private static final int N = 5001;

public static void solve() {
    int n = io.nextInt();
    int[] cnt = new int[N];
    for (int i = 0; i < n; i++) {
        int x = io.nextInt();
        if (x < n) cnt[x]++;
    }

    int mex = 0;
    while (cnt[mex] > 0) mex++;
    long[] dp = new long[mex + 1];
    Arrays.fill(dp, Long.MAX_VALUE);
    dp[mex] = 0;
    for (int i = mex - 1; i >= 0; i--) {
        for (int j = i + 1; j <= mex; j++) {
            dp[i] = Math.min(dp[i], dp[j] + (long) (cnt[i] - 1) * j + i);
        }
    }
    io.println(dp[0]);
}

2023-10-02发表2023-10-02更新算法 / AtCoder4 分钟读完 (大约663个字)

AtCoder Beginner Contest 322

First ABC 2

public static void solve() {
    int n = io.nextInt();
    String s = io.next();
    int ans = s.indexOf("ABC");
    io.println(ans < 0 ? -1 : ans + 1);
}

Prefix and Suffix

public static void solve() {
    int n = io.nextInt(), m = io.nextInt();
    String s = io.next(), t = io.next();
    int ans = 0;
    for (int i = 0; i < n; i++) {
        if (s.charAt(i) != t.charAt(i)) {
            ans |= 2;
        }
        if (s.charAt(i) != t.charAt(m - n + i)) {
            ans |= 1;
        }
    }
    io.println(ans);
}

Festival

public static void solve() {
    int n = io.nextInt(), m = io.nextInt();
    int[] ans = new int[n];
    Arrays.fill(ans, -1);
    for (int i = 0; i < m; i++) {
        int x = io.nextInt() - 1;
        ans[x] = 0;
    }
    for (int i = n - 2; i >= 0; i--) {
        if (ans[i] == -1) {
            ans[i] = ans[i + 1] + 1;
        }
    }
    for (int i = 0; i < n; i++) {
        io.println(ans[i]);
    }
}

Polyomino

模拟题，使用位运算似乎更简单，题解。

public static void solve() {
    char[][][] G = new char[3][4][4];
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            G[i][j] = io.next().toCharArray();
        }
    }
    io.println(dfs(0, G, new char[4][4]) ? "Yes" : "No");
}

private static boolean dfs(int i, char[][][] G, char[][] C) {
    if (C == null) return false;
    if (i == 3) return check(C);
    for (int j = 0; j < 4; j++) {
        for (int dx = -3; dx <= 3; dx++) {
            for (int dy = -3; dy <= 3; dy++) {
                char[][] T = move(dx, dy, G[i]);
                if (T == null) continue;
                if (dfs(i + 1, G, add(T, C))) return true;
            }
        }
        G[i] = rotate(G[i]);
    }
    return false;
}

private static char[][] rotate(char[][] A) {
    char[][] B = new char[4][4];
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            B[i][j] = A[3 - j][i];
        }
    }
    return B;
}

private static char[][] move(int dx, int dy, char[][] G) {
    char[][] res = new char[4][4];
    for (int i = 0; i < 4; i++) {
        Arrays.fill(res[i], '.');
    }
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            int nx = i + dx, ny = j + dy;
            if(G[i][j] == '#') {
                if (nx < 0 || nx >= 4 || ny < 0 || ny >= 4) {
                    return null;
                }
                res[nx][ny] = G[i][j];
            }
        }
    }
    return res;
}

private static char[][] add(char[][] T, char[][] C) {
    char[][] res = new char[4][4];
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            if (C[i][j] == '#' && T[i][j] == '#') return null;
            if (C[i][j] == '#' || T[i][j] == '#') res[i][j] = '#';
            else res[i][j] = '.';
        }
    }
    return res;
}

private static boolean check(char[][] C) {
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            if (C[i][j] != '#') return false;
        }
    }
    return true;
}

Product Development

动态规划，\(dp[i][j]\) 表示从前 \(i\) 个计划中选择开发计划，使得参数到达 \(j\)，需要的最小成本，其中 \(j\) 表示 \(a_{1},a_{2},\dots,a_{k}\) 的某个取值，通过将序列看作 \(p+1\) 进制数，可以将一个序列转换为某个数值（在这里就是 \(j\)）。

public static void solve() {
    int n = io.nextInt(), k = io.nextInt(), p = io.nextInt();

    int[] pw = new int[k + 1];
    pw[0] = 1;
    for (int i = 1; i <= k; i++) {
        pw[i] = pw[i - 1] * (p + 1);
    }

    long[] dp = new long[pw[k]];
    Arrays.fill(dp, -1L);
    dp[0] = 0;

    for (int i = 0; i < n; i++) {
        int c = io.nextInt();
        int[] a = new int[k];
        for (int j = 0; j < k; j++) {
            a[j] = io.nextInt();
        }

        for (int s = pw[k] - 1; s >= 0; s--) {
            int t = 0;
            for (int j = 0; j < k; j++) {
                int cur = s / pw[j] % (p + 1);
                int nxt = Math.min(p, cur + a[j]);
                t += nxt * pw[j];
            }
            if (dp[s] != -1 && (dp[t] == -1 || dp[t] > dp[s] + c)) {
                dp[t] = dp[s] + c;
            }
        }
    }
    io.println(dp[pw[k] - 1]);
}

有序三元组中的最大值 I