Case05_区间覆盖问题，POJ2376题解

Case05_区间覆盖问题

区间覆盖问题贪心策略：
首先对区间端点s排序
（case03里有相关解释）
设置游标 start end
初始化都为1
s<=start
区间可能覆盖start
比较end和t大小，始终保持end是最大的
s>start
计数加1
更新start为end+1
并判断此时的s<=start 是 比较end和t 否 终止此次循环，进行下次循环

比较完s
看t>end
终止循环

比较绕
模板题

POJ2376

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.
Input
* Line 1: Two space-separated integers: N and T

* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.
Output
* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.
Sample Input
3 10
1 7
3 6
6 10
Sample Output
2
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

INPUT DETAILS:

There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10.

OUTPUT DETAILS:

By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

代码如下

package greed_dynamic;

import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.max;

public class Case05_区间覆盖问题 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int T = sc.nextInt();
		Job[] jobs = new Job[N];
		for (int i = 0; i < N; i++) {
			jobs [i] = new Job(sc.nextInt(),sc.nextInt());
		}
		Arrays.sort(jobs);
		int start = 1;
		int end = 1;
		int ans = 1;
		
		for (int i = 0; i < N; i++) {
			int s = jobs[i].s;
			int t = jobs[i].t;
			
			if (i == 0 && s > 1) {
				break;
			}
			if (s<=start) {
				end = max(t,end);
			} else {//开始下一个区间
				ans++;
				start = end + 1;//更新起点
				if (s <= start) {//当前区间有可能覆盖start
					end = max(t,end);
				} else {
					break;
				}
			}
			if (end >= T) { //当前的end超越了线段的右侧
				break;
			}
		}
		if (end < T) {
			System.out.println(-1);
		} else {
			System.out.println(ans);
		}
	}
	public static class Job implements Comparable<Job> {
        int s;
        int t;
        
        public Job(int s,int t) {
        	this.s = s;
        	this.t = t;
        }
		@Override
		public int compareTo(Job other) {
			// TODO 自动生成的方法存根
			int x = this.s - other.s;
			if (x == 0) {
				return this.t - this.t;
			}
			return x;
		} 
		
	}

}