Code Jam Kickstart 2019 Round A 題解

Problem A

題目

題意：

分析

直接滑動窗口解決O(n)

代碼

class Solution {
public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int t = in.nextInt();
        for (int num = 1; num <= t; ++num) {
            int N = in.nextInt();
            int P = in.nextInt();
            int[] S = new int[N];
            for (int j = 0; j < N; j++) {
                S[j] = in.nextInt();
            }

            int res = Integer.MAX_VALUE;
            Arrays.sort(S);
            int left = 0, right = P - 1;
            int sum = 0;
            for (int i = left; i <= right; i++) sum += S[i];
            res = Math.min(res, P * S[right] - sum);
            while (right < N) {
                sum -= S[left];
                left++;
                right++;
                if (right == N) break;
                sum += S[right];
                res = Math.min(res, P * S[right] - sum);
            }
            System.out.println("Case #" + num + ": " + res);
        }
    }
}

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Problem B

題目

題意：

分析

• 多源BFS + 二分

  • (1) 先使用多源BFS，從多個郵局出發，求出每一個點到郵局的最短距離，複雜度是O(RC，而後獲得了最大的距離，
    而且可以獲得最大的距離max_dist即overall delivery time

  • (2) 而後使用二分搜索，可否找到一個K，在0-max_dist之間，知足若是在某處新建一個郵局，可否使得最大的delivery time（即最大的距離）至可能是K
    複雜度O(RClog(R+C))
    也就是說 對於一個固定的K, 咱們判斷，是否能夠找到一個點放置郵局使得，全部點到delivers的距離不會超過K.
    這裏判斷的時候還用到了個數據小技巧，就是曼哈頓距離與切比雪夫距離的轉化:
    dist((x1, y1), (x2, y2)) = max(abs(x1 + y1 - (x2 + y2)), abs(x1 - y1 - (x2 - y2)))

代碼

public class Solution {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int t = in.nextInt();
        for (int num = 1; num <= t; ++num) {
            int R = in.nextInt();
            int C = in.nextInt();
            String[] grid = new String[R];
            for (int i = 0; i < R; i++) {
                grid[i] = in.next();
            }
            char[][] graph = new char[R][C];
            for (int i = 0; i < R; i++) {
                for (int j = 0; j < C; j++) {
                    graph[i][j] = grid[i].charAt(j);
                }
            }
            int res = solveLarge(graph, R, C);
            System.out.println("Case #" + num + ": " + res);
        }
    }

    static int[][] dist;
    static int[][] dirs = {{1,0}, {0, 1}, {-1, 0}, {0, -1}};

    private static int solveLarge(char[][] graph, int m, int n) {
        int res = 0;
        dist = new int[m][n];
        int hi = multiBFS(graph, m, n);
        int low = 0;
        while (low <= hi) {
            int mid = (low + hi) >> 1;
            if (isValid(mid, m, n)) hi = mid - 1;
            else low = mid + 1;
        }
        res = hi + 1;
        return res;
    }

    private static int multiBFS(char[][] graph, int m, int n) {
        int maxDist = 0;
        Queue<int[]> queue = new LinkedList<>();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dist[i][j] = -1;
                if (graph[i][j] == '1') {
                    queue.add(new int[]{i, j});
                    dist[i][j] = 0;
                }
            }
        }
        while (!queue.isEmpty()) {
            int[] cur = queue.poll();
            maxDist = dist[cur[0]][cur[1]];
            for (int[] d : dirs) {
                int nx = cur[0] + d[0];
                int ny = cur[1] + d[1];
                if (nx < 0 || nx >= m || ny < 0 || ny >= n || dist[nx][ny] != -1) continue;
                dist[nx][ny] = maxDist + 1;
                queue.add(new int[]{nx, ny});
            }
        }
        return maxDist;
    }

    private static boolean isValid(int mid, int r, int c) {
        int x = Integer.MIN_VALUE;
        int y = Integer.MAX_VALUE;
        int z = Integer.MIN_VALUE;
        int w = Integer.MAX_VALUE;
        for(int i = 0; i < r ; ++i)
            for(int j = 0 ; j < c ; ++j)
                if(dist[i][j] > mid){
                    x = Math.max(x, i+j);
                    y = Math.min(y,i+j);
                    z = Math.max(z,i-j);
                    w = Math.min(w,i-j);
                }
        if(w == Integer.MAX_VALUE) return true;
        for(int i = 0 ; i < r; ++i)
            for(int j = 0 ; j < c; ++j)
                if(Math.abs(x - (i+j)) <= mid &&
                        Math.abs(y - (i+j)) <= mid &&
                        Math.abs(z - (i-j)) <= mid &&
                        Math.abs(w - (i-j)) <= mid)
                    return true;
        return false;
    }
}

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Problem C

題目

題意：

分析

代碼