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package 第02期.mca_02;
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import java.util.Arrays;
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// 每周有营养的大厂算法面试题
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// 课时92~课时93
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// 来自字节飞书团队
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// 在字节跳动，大家都使用飞书的日历功能进行会议室的预订，遇到会议高峰时期，
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// 会议室就可能不够用，现在请你实现一个算法，判断预订会议时是否有空的会议室可用。
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// 为简化问题，这里忽略会议室的大小，认为所有的会议室都是等价的，
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// 只要空闲就可以容纳任意的会议，并且：
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// 1. 所有的会议预订都是当日预订当日的时段
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// 2. 会议时段是一个左闭右开的时间区间，精确到分钟
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// 3. 每个会议室刚开始都是空闲状态，同一时间一个会议室只能进行一场会议
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// 4. 会议一旦预订成功就会按时进行
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// 比如上午11点到中午12点的会议即[660, 720)
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// 给定一个会议室总数m
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// 一个预定事件由[a,b,c]代表 :
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// a代表预定动作的发生时间，早来早得; b代表会议的召开时间; c代表会议的结束时间
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// 给定一个n*3的二维数组，即可表示所有预定事件
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// 返回一个长度为n的boolean类型的数组，表示每一个预定时间是否成功
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public class Code06_MeetingCheck {
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public static boolean[] reserveMeetings(int m, int[][] meetings) {
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// 会议的总场次
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int n = meetings.length;
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// 开头时间，结尾时间
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int[] ranks = new int[n << 1];
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for (int i = 0; i < n; i++) {
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ranks[i] = meetings[i][1];
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ranks[i + n] = meetings[i][2] - 1;
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}
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Arrays.sort(ranks);
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// 0 : [6, 100, 200]
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// 1 : [4, 30, 300]
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// 30,1 100,2 200,3 300,4
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// [0,6,2,3]
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// [1,4,1,4]
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//
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// 0 T/F , 1, T/ 2,
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// [1,4,1,4] [0,6,2,3] ....
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int[][] reMeetings = new int[n][4];
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int max = 0;
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for (int i = 0; i < n; i++) {
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reMeetings[i][0] = i;
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reMeetings[i][1] = meetings[i][0];
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reMeetings[i][2] = rank(ranks, meetings[i][1]);
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reMeetings[i][3] = rank(ranks, meetings[i][2] - 1);
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max = Math.max(max, reMeetings[i][3]);
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}
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SegmentTree st = new SegmentTree(max);
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Arrays.sort(reMeetings, (a, b) -> a[1] - b[1]);
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boolean[] ans = new boolean[n];
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for (int[] meeting : reMeetings) {
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if (st.queryMax(meeting[2], meeting[3]) < m) {
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ans[meeting[0]] = true;
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st.add(meeting[2], meeting[3], 1);
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}
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}
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return ans;
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}
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// 返回>=num, 最左位置
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public static int rank(int[] sorted, int num) {
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int l = 0;
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int r = sorted.length - 1;
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int m = 0;
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int ans = 0;
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while (l <= r) {
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m = (l + r) / 2;
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if (sorted[m] >= num) {
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ans = m;
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r = m - 1;
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} else {
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l = m + 1;
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}
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}
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return ans + 1;
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}
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public static class SegmentTree {
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private int n;
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private int[] max;
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private int[] lazy;
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public SegmentTree(int maxSize) {
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n = maxSize;
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max = new int[n << 2];
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lazy = new int[n << 2];
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}
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private void pushUp(int rt) {
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max[rt] = Math.max(max[rt << 1], max[rt << 1 | 1]);
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}
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private void pushDown(int rt, int ln, int rn) {
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if (lazy[rt] != 0) {
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lazy[rt << 1] += lazy[rt];
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max[rt << 1] += lazy[rt];
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lazy[rt << 1 | 1] += lazy[rt];
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max[rt << 1 | 1] += lazy[rt];
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lazy[rt] = 0;
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}
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}
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public void add(int L, int R, int C) {
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add(L, R, C, 1, n, 1);
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}
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private void add(int L, int R, int C, int l, int r, int rt) {
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if (L <= l && r <= R) {
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max[rt] += C;
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lazy[rt] += C;
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return;
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}
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int mid = (l + r) >> 1;
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pushDown(rt, mid - l + 1, r - mid);
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if (L <= mid) {
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add(L, R, C, l, mid, rt << 1);
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}
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if (R > mid) {
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add(L, R, C, mid + 1, r, rt << 1 | 1);
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}
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pushUp(rt);
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}
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public int queryMax(int L, int R) {
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return queryMax(L, R, 1, n, 1);
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}
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private int queryMax(int L, int R, int l, int r, int rt) {
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if (L <= l && r <= R) {
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return max[rt];
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}
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int mid = (l + r) >> 1;
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pushDown(rt, mid - l + 1, r - mid);
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int ans = 0;
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if (L <= mid) {
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ans = Math.max(ans, queryMax(L, R, l, mid, rt << 1));
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}
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if (R > mid) {
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ans = Math.max(ans, queryMax(L, R, mid + 1, r, rt << 1 | 1));
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}
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return ans;
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}
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}
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// 为了测试线段树
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public static class Right {
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public int[] arr;
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public Right(int maxSize) {
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arr = new int[maxSize + 1];
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}
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public void add(int L, int R, int C) {
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for (int i = L; i <= R; i++) {
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arr[i] += C;
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}
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}
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public int queryMax(int L, int R) {
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int ans = 0;
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for (int i = L; i <= R; i++) {
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ans = Math.max(ans, arr[i]);
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}
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return ans;
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}
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}
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// 测试线段树的对数器
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public static void main(String[] args) {
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int N = 50;
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int V = 10;
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int testTimes1 = 1000;
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int testTimes2 = 1000;
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System.out.println("测试线段树开始");
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for (int i = 0; i < testTimes1; i++) {
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int n = (int) (Math.random() * N) + 1;
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SegmentTree st = new SegmentTree(n);
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Right right = new Right(n);
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for (int j = 0; j < testTimes2; j++) {
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int a = (int) (Math.random() * n) + 1;
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int b = (int) (Math.random() * n) + 1;
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int L = Math.min(a, b);
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int R = Math.max(a, b);
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if (Math.random() < 0.5) {
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int C = (int) (Math.random() * V);
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st.add(L, R, C);
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right.add(L, R, C);
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} else {
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if (st.queryMax(L, R) != right.queryMax(L, R)) {
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System.out.println("出错了!");
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}
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}
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}
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}
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System.out.println("测试线段树结束");
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}
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}
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algorithmzuo
2 years ago