45D--Event Dates

题目

On a history lesson the teacher asked Vasya to name the dates when n famous events took place. He doesn’t remembers the exact dates but he remembers a segment of days [li, ri] (inclusive) on which the event could have taken place. However Vasya also remembers that there was at most one event in one day. Help him choose such n dates of famous events that will fulfill both conditions. It is guaranteed that it is possible.

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of known events. Then follow n lines containing two integers li and ri each (1 ≤ li ≤ ri ≤ 107) — the earliest acceptable date and the latest acceptable date of the i-th event.

Output

Print n numbers — the dates on which the events took place. If there are several solutions, print any of them. It is guaranteed that a solution exists.

Examples

input

3
1 2
2 3
3 4

output

1 2 3

input

2
1 3
1 3

output

1 2

题意：老师询问 Vasya历史事件发生的日期，但是 Vasya不记得事件发生的具体事件，但是记得该事件发生的某个事件段，但是每天只能发生一件事件，请你帮组 Vasya算n个事件有可能发生的事件。

解题思路：现将每个事件的发生事件段记录在list集合中，然后按照最迟事件对其进行排序 ，建立一个hashSet集合，再遍历每个事件是时间段，将不存在HashSet的某个时间存入该Set集合中（仅保存其中一个）。并将该时间存入list集合中，然后在将其按照事件序号排序，最后遍历list集合输出每个事件可能发生的时间。

AC–Code

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.Scanner;
public class CF45D {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		ArrayList<Date> list = new ArrayList<Date>();
		HashSet<Integer> set = new HashSet<Integer>();
		for(int i =0;i<n;i++)
		{
			Date obj = new Date();
			obj.setL(sc.nextInt());
			obj.setR(sc.nextInt());
			obj.setI(i);
			obj.setAns(0);
			list.add(obj);
		}
		Collections.sort(list);
		for (Date date : list) {
			for(int i=date.getL();i<=date.getR();i++) {
				if(!set.contains(i))
				{
					date.setAns(i);
					set.add(i);
					break;
				}
			}
		}
		Collections.sort(list);
		for (Date date : list) {
			System.out.print(date.getAns()+" ");
		}
		sc.close();
	}

}
class Date implements Comparable<Date>{
	private int l,r,i,ans;

	public int getL() {
		return l;
	}

	public void setL(int l) {
		this.l = l;
	}

	public int getR() {
		return r;
	}

	public void setR(int r) {
		this.r = r;
	}

	public int getI() {
		return i;
	}

	public void setI(int i) {
		this.i = i;
	}

	public int getAns() {
		return ans;
	}

	public void setAns(int ans) {
		this.ans = ans;
	}
	public int compareTo(Date o) {
		if(o.ans==0)
		{
			if (this.r < o.r) {
	            return -1;
	        } else if (this.r > o.r) {
	            return 1;
	        } else {
	            return 0;
	        }
		}
		else {
			if (this.i < o.i) {
	            return -1;
	        } else if (this.i > o.i) {
	            return 1;
	        } else {
	            return 0;
	        }
		}
	}
}

学到的知识：HashSet中保存的数据是唯一不重复的数据，在构建list集合时，先对其类中的ans赋值为0，方便后续进行不同方法的排序。在第一次排序中，所有事件的发生的时间设置为0，则按照最迟事件排序，第二次排序，已经知道了每个事件的发生时间（ans！=0），然后对其按照事件序号排序。