Java实现蓝桥杯 算法提高 线段和点

算法提高 线段和点
时间限制:1.0s 内存限制:256.0MB
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问题描述
  有n个点和m个区间,点和区间的端点全部是整数,对于点a和区间[b,c],若a>=b且a<=c,称点a满足区间[b,c]。
  求最小的点的子集,使得所有区间都被满足。
输入格式
  第一行两个整数n m
  以下n行 每行一个整数,代表点的坐标
  以下m行 每行两个整数,代表区间的范围
输出格式
  输出一行,最少的满足所有区间的点数,如无解输出-1。
样例输入
5 5
2
6
3
8
7
2 5
3 4
3 3
2 7
6 9
样例输出
2
数据规模和约定
  1<=n,m<=10000
  0<=点和区间的坐标<=50000

import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.Comparator;


public class xianduanhedian {
	private static InputStream is = System.in;

	public static int nextInt() {
		try {
			int i;

			while ((i = is.read()) < 45 || i > 57) {
			}

			int mark = 1, temp = 0;

			if (i == 45) {
				mark = -1;
				i = is.read();
			}

			while (i > 47 && i < 58) {
				temp = temp * 10 + i - 48;
				i = is.read();
			}

			return temp * mark;
		} catch (IOException e) {
			e.printStackTrace();
		}

		return -1;
	}

	static class Node {
		public int start;
		public int end;

		public Node(int start, int end) {
			this.start = start;
			this.end = end;
		}

	}

	public static void main(String[] args) {
		int n = nextInt();
		int m = nextInt();
		int point[] = new int[n];
		for (int i = 0; i < n; i++)
			point[i] = nextInt();
		Node node[] = new Node[m];
		for (int i = 0; i < m; i++)
			node[i] = new Node(nextInt(), nextInt());
		Arrays.sort(point);
		Arrays.sort(node, new Comparator<Node>() {
			public int compare(Node o1, Node o2) {
				return o1.end - o2.end;
			}
		});
		int currentPoint = 0;
		int count = 0;
		int j = 1;
		for (int i = 0; i < m; i++) {
			int x = node[i].start;
			int y = node[i].end;
			if (x <= currentPoint)
				continue;
			int temp = -1;
			for (j -= 1; j < n; j++) {
				if (point[j] <= y) {
					temp = point[j];
				} else {
					break;
				}
			}
			if (temp == -1) {
				count = 0;
				break;
			} else {
				currentPoint = temp;
				count++;
			}

		}
		System.out.println(count);
	}

}

原文地址:https://www.cnblogs.com/a1439775520/p/13079240.html