美团

/**
 * 输入数组的元素只能为 1或-1
 * 求解乘积为1的子串
 * 4
 * 1 1 -1 -1
 * 10
 *     // 前缀积
 *     public static void main(String[] args) {
 *         Scanner sc = new Scanner(System.in);
 *         int n = sc.nextInt();
 *         int[] nums = new int[n];
 *         for (int i = 0; i < n; i++) nums[i] = sc.nextInt();
 *         // 前缀积 [0, i]
 *         for (int i = 1; i < n; i++) nums[i] *= nums[i-1];
 *
 *         int ans = 0;
 *         for (int i = 0; i < n; i++) {
 *             for (int j = i; j < n; j++) {
 *                 if (nums[j] * (i > 0? nums[i-1]: 1) == 1) ans ++;
 *             }
 *         }
 *         System.out.println(ans);
 *     }
 *
 */

/**
 * n个房间 m秒
 * 接下来m个数字代表第i秒 炸弹在哪个房间爆炸
 * dp[i][j]表示第i秒在第j个房间所消耗的能量
 * 2 4
 * 2 1 1 2
 * 2
 *
 * 3 10
 * 2 3 1 3 2 1 1 2 3 1
 *
 * static int inf = (int)(1e9+5);
 *     public static void main(String[] args) {
 *         Scanner sc = new Scanner(System.in);
 *         int n = sc.nextInt();
 *         int m = sc.nextInt();
 *         int[] arr = new int[m+1];
 *         for (int i = 1; i <= m; i++) arr[i] = sc.nextInt();
 *         int[][] dp = new int[m+1][n+1];
 *         for (int i = 0; i <= m; i++) Arrays.fill(dp[i], inf);
 *         dp[0][1] = 0;
 *         for (int i = 1; i <= m; i++) {
 *             for (int j = 1; j <= n; j++) {
 *                 if (arr[i] == j) dp[i][j] = inf;
 *                 else dp[i][j] = Math.min(dp[i][j], Math.min(dp[i-1][j], dp[i-1][arr[i]] + 1));    // 不跳 or 跳
 *             }
 *         }
 *         Arrays.sort(dp[m]);
 *         System.out.println(dp[m][0]);
 *     }
 */

/**
 * 节点范围n, 1 <=n<= 10000
 * 邻接表建图
 * 6
 * 1 0 1 1 0 0 0为白色节点 1为黑色节点
 * 0 1 2 1 4 4
 * 3 2
 * 若白色节点 没有子节点或者至少有一个黑色节点 则为好节点
 * 若黑色节点 没有子节点或者所有节点全为白色节点 则为好节点
 */
public class Main {
    static int N = 10005, M = 2 * 10005;
    static int[] head = new int[N];
    static int[] edge = new int[M];
    static int[] next = new int[M];

    static int root, r1, r2;
    static int idx = 0;
    static void add(int x, int y) {    // x -> y 头插法
        edge[idx] = y;
        next[idx] = head[x];
        head[x] = idx++;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] color = new int[n+1];
        for (int i = 1; i <= n; i++) color[i] = sc.nextInt();
        Arrays.fill(head, -1);
        for (int i = 1; i <= n; i++) {
            int from = sc.nextInt();
            if (from == 0) root = i;
            else add(from ,i);
        }

        dfs(color, root);
        System.out.println(r1 + " " + r2);
    }

    public static void dfs(int[] color, int node) {
        // 处理当前节点
        if (color[node] == 0) { // 如果是白色节点
            int size = 0, black = 0;
            for (int i = head[node]; i != -1; i = next[i]) {
                size++;
                int to = edge[i];
                if (color[to] == 1) black++;
            }
            if (size == 0 || black >= 1) r1++;
        } else {    // 如果为黑色节点
            int size = 0, white = 0;
            for (int i = head[node]; i != -1; i = next[i]) {
                size ++;
                int to = edge[i];
                if (color[to] == 0) white++;
            }
            if (size == 0 || white == size) r2++;
        }
        // 处理子节点
        for (int i = head[node]; i != -1; i = next[i]) {
            int to = edge[i];
            dfs(color, to);
        }
    }
}

 

// https://www.lintcode.com/problem/386/description

public class Solution {
    /**
     * @param s: A string
     * @param k: An integer
     * @return: An integer
     */
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        // write your code here
        // 使用Map记录(字符，出现次数)
        Map<Character, Integer> map = new HashMap<>();
        int sx = 0, sy = 0;
        int left = 0, right = 0;
        int maxLen = 0;
        for (left = 0; left < s.length(); left ++) {
            while (right < s.length()) {
                char c = s.charAt(right);
                if (map.containsKey(c)) {
                    map.put(c, map.get(c) + 1);
                } else {
                    if (map.size() == k) {
                        break;
                    } else {
                        map.put(c, 1);
                    }
                }
                right ++;
            }


            if (maxLen < right - left) {
                maxLen = right - left;
                sx = left;
                sy = right;
            }
            // maxLen = Math.max(maxLen, right - left);



            // left左移
            char c = s.charAt(left);
            if (map.containsKey(c)) {
                if (map.get(c) == 1) {
                    map.remove(c);
                } else {
                    map.put(c, map.get(c) - 1);
                }
            }
        }
        System.out.println(s.substring(sx, sy));　　// 也可以把结果打印一下
        return maxLen;
    }
}