反向考虑，以所有可以直接到达的点作为起点。

若是 i → j，i j 处的值奇偶不同，那么 i 的结果就为1，i 是起点。

若是 i → j，i j 处的值奇偶相同，那么就以 j → i 为边建图。

public static void solve(Kattio kattio) {
        int n = kattio.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = kattio.nextInt();
        }
        int[] res = new int[n];
        List<Integer>[] graph = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            graph[i] = new ArrayList<>();
        }
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            int prev = i - a[i];
            int next = i + a[i];
            int odd = a[i] % 2;
            if ((prev >= 0 && a[prev] % 2 != odd) || (next < n && a[next] % 2 != odd)) {
                res[i] = 1;
                queue.offer(i);
            } else {
                res[i] = -1;
            }
            if (prev >= 0 && a[prev] % 2 == odd) {
                graph[prev].add(i);
            }
            if (next < n && a[next] % 2 == odd) {
                graph[next].add(i);
            }
        }
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int id = queue.poll();
                for (Integer next : graph[id]) {
                    if (res[next] < 0) {
                        res[next] = res[id] + 1;
                        queue.offer(next);
                    }
                }
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int re : res) {
            sb.append(re).append(" ");
        }
        kattio.println(sb);
    }