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package class_2023_04_3_week;
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// 来自学员问题，蓝桥杯练习题
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// 给定一个长度为n的数组arr
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// 现在你有一次机会, 将其中连续的K个数全修改成任意一个值
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// 请你计算如何修改可以使修改后的数 列的最长不下降子序列最长
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// 请输出这个最长的长度。
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// 最长不下降子序列:子序列中的每个数不小于在它之前的数
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// 1 <= k, n <= 10^5
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// 1 <= arr[i] <= 10^6
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// 测试链接 : https://www.luogu.com.cn/problem/P8776
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下的所有代码，并把主类名改成"Main"
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// 可以直接通过
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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public class Code03_LongestNoDecreaseModifyKSubarray {
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public static int MAXN = 100001;
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public static int[] arr = new int[MAXN];
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public static int[] right = new int[MAXN];
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public static int[] ends = new int[MAXN];
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public static int n, k;
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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n = (int) in.nval;
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in.nextToken();
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k = (int) (in.nval);
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for (int i = 1; i <= n; i++) {
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in.nextToken();
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arr[i] = (int) in.nval;
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}
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if (k >= n) {
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out.println(n);
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} else {
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right();
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out.println(getAns());
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}
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out.flush();
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}
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}
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public static void right() {
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right[n] = 1;
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ends[1] = arr[n];
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int len = 1;
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for (int i = n - 1; i > 0; i--) {
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int l = 1;
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int r = len;
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int m, find = len + 1;
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while (l <= r) {
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m = (l + r) / 2;
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if (ends[m] < arr[i]) {
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find = 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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ends[find] = arr[i];
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len = Math.max(len, find);
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right[i] = find;
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}
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}
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public static int getAns() {
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int ans = 0;
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int len = 0;
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for (int i = k + 1, j = 1; i <= n; i++, j++) {
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int l = 1;
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int r = len;
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int m, find = len + 1;
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while (l <= r) {
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m = (l + r) / 2;
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// 当前的arr[i], 利用之前的ends，求左边长度！
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if (ends[m] > arr[i]) {
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find = 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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ans = Math.max(ans, find + right[i] - 1 + k);
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l = 1;
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r = len;
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find = len + 1;
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while (l <= r) {
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m = (l + r) / 2;
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if (ends[m] > arr[j]) {
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find = 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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len = Math.max(len, find);
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ends[find] = arr[j];
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}
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ans = Math.max(ans, len + k);
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return ans;
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}
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} |
