题目1
P3834 【模板】可持久化线段树 2 - 洛谷
// 区间内第k小，第一种写法，java版 // 给定一个长度为n的数组，接下来有m条查询，格式如下 // 查询 l r k : 打印[l..r]范围内第k小的值 // 1 <= n、m <= 2 * 10^5 // 1 <= 数组中的数字 <= 10^9 // 测试链接 : https://www.luogu.com.cn/problem/P3834 // 本题是讲解157，可持久化线段树模版题，现在作为整体二分的模版题 // 提交以下的code，提交时请把类名改成"Main"，可以通过所有测试用例 import java.io.IOException; import java.io.InputStream; import java.io.OutputStreamWriter; import java.io.PrintWriter; import java.util.Arrays; public class Code01_RangeKth1 { public static int MAXN = 200001; public static int n, m; // 位置i，数值v public static int[][] arr = new int[MAXN][2]; // 查询 public static int[] qid = new int[MAXN]; public static int[] l = new int[MAXN]; public static int[] r = new int[MAXN]; public static int[] k = new int[MAXN]; // 树状数组 public static int[] tree = new int[MAXN]; // 整体二分 public static int[] lset = new int[MAXN]; public static int[] rset = new int[MAXN]; // 查询的答案 public static int[] ans = new int[MAXN]; // 树状数组中的lowbit public static int lowbit(int i) { return i & -i; } // 树状数组中增加i位置的词频 public static void add(int i, int v) { while (i <= n) { tree[i] += v; i += lowbit(i); } } // 树状数组中查询[1~i]范围的词频累加和 public static int sum(int i) { int ret = 0; while (i > 0) { ret += tree[i]; i -= lowbit(i); } return ret; } // 树状数组中查询[l~r]范围的词频累加和 public static int query(int l, int r) { return sum(r) - sum(l - 1); } // 整体二分的第一种写法 // 问题范围[ql..qr]，答案范围[vl..vr]，答案范围的每个下标都是数字的排名 public static void compute(int ql, int qr, int vl, int vr) { if (ql > qr) { return; } if (vl == vr) { for (int i = ql; i <= qr; i++) { ans[qid[i]] = arr[vl][1]; } } else { // 修改数据状况 int mid = (vl + vr) / 2; for (int i = vl; i <= mid; i++) { add(arr[i][0], 1); } // 检查每个问题并划分左右 int lsiz = 0, rsiz = 0; for (int i = ql; i <= qr; i++) { int id = qid[i]; int satisfy = query(l[id], r[id]); if (satisfy >= k[id]) { lset[++lsiz] = id; } else { k[id] -= satisfy; rset[++rsiz] = id; } } for (int i = 1; i <= lsiz; i++) { qid[ql + i - 1] = lset[i]; } for (int i = 1; i <= rsiz; i++) { qid[ql + lsiz + i - 1] = rset[i]; } // 撤回数据状况 for (int i = vl; i <= mid; i++) { add(arr[i][0], -1); } // 左右两侧各自递归 compute(ql, ql + lsiz - 1, vl, mid); compute(ql + lsiz, qr, mid + 1, vr); } } public static void main(String[] args) throws Exception { FastReader in = new FastReader(System.in); PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out)); n = in.nextInt(); m = in.nextInt(); for (int i = 1; i <= n; i++) { arr[i][0] = i; arr[i][1] = in.nextInt(); } for (int i = 1; i <= m; i++) { qid[i] = i; l[i] = in.nextInt(); r[i] = in.nextInt(); k[i] = in.nextInt(); } Arrays.sort(arr, 1, n + 1, (a, b) -> a[1] - b[1]); compute(1, m, 1, n); for (int i = 1; i <= m; i++) { out.println(ans[i]); } out.flush(); out.close(); })
