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package class_2022_08_2_week;
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// 给定平面上n个点，x和y坐标都是整数
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// 找出其中的一对点的距离，使得在这n个点的所有点对中，该距离为所有点对中最小的
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// 返回最短距离，精确到小数点后面4位
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// 测试链接 : https://www.luogu.com.cn/problem/P1429
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// 提交以下所有代码，把主类名改成Main，可以直接通过
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// T(N) = 2*T(N/2) + O(N)
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// 这个表达式我们很熟悉，和归并排序一样的表达式
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// 时间复杂度是O(N*logN)
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// 需要用到归并排序的技巧才能做到
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// 我们课上的独家
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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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import java.util.Arrays;
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public class Code04_ClosestTwoPoints2 {
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public static int N = 200001;
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public static Point[] points = new Point[N];
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public static Point[] merge = new Point[N];
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public static Point[] deals = new Point[N];
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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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int n = (int) in.nval;
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for (int i = 0; i < n; i++) {
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in.nextToken();
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double x = (double) in.nval;
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in.nextToken();
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double y = (double) in.nval;
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points[i] = new Point(x, y);
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}
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Arrays.sort(points, 0, n, (a, b) -> a.x <= b.x ? -1 : 1);
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double ans = nearest(0, n - 1);
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out.println(String.format("%.4f", ans));
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out.flush();
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}
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}
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public static class Point {
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public double x;
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public double y;
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public Point(double a, double b) {
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x = a;
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y = b;
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}
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}
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public static double nearest(int left, int right) {
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double ans = Double.MAX_VALUE;
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if (left == right) {
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return ans;
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}
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int mid = (right + left) / 2;
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double midX = points[mid].x;
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ans = Math.min(nearest(left, mid), nearest(mid + 1, right));
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int p1 = left;
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int p2 = mid + 1;
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int mergeSize = left;
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int dealSize = 0;
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while (p1 <= mid && p2 <= right) {
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merge[mergeSize] = points[p1].y <= points[p2].y ? points[p1++] : points[p2++];
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if (Math.abs(merge[mergeSize].x - midX) <= ans) {
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deals[dealSize++] = merge[mergeSize];
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}
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mergeSize++;
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}
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while (p1 <= mid) {
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merge[mergeSize] = points[p1++];
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if (Math.abs(merge[mergeSize].x - midX) <= ans) {
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deals[dealSize++] = merge[mergeSize];
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}
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mergeSize++;
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}
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while (p2 <= right) {
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merge[mergeSize] = points[p2++];
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if (Math.abs(merge[mergeSize].x - midX) <= ans) {
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deals[dealSize++] = merge[mergeSize];
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}
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mergeSize++;
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}
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for (int i = left; i <= right; i++) {
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points[i] = merge[i];
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}
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for (int i = 0; i < dealSize; i++) {
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for (int j = i + 1; j < dealSize; j++) {
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if (deals[j].y - deals[i].y >= ans) {
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break;
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}
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ans = Math.min(ans, distance(deals[i], deals[j]));
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}
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}
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return ans;
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}
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public static double distance(Point a, Point b) {
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double x = a.x - b.x;
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double y = a.y - b.y;
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return Math.sqrt(x * x + y * y);
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}
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} |
