宣传栏清理问题

问题描述：

在学校的学生公寓楼内，每栋楼大厅内的宣传栏上每天都有大量的海报。

为了使用一种特殊的方法清理宣传栏而不留下难看的痕迹，事先需要知道宣传栏上张贴

海报的位置信息。

为了简化问题我们规定：

1. 宣传栏用一个区间[0, 1000000]来表示。

2. 假设海报的高度均与宣传栏的高度相同，各海报用两个整数left 和

right（right>left）来表示，表示海报占据区间[left，right]。

3. 两张海报是可以部分或者全部重叠的。

4. 所有海报都不会超过宣传栏的大小。

根据给定的每张海报所占据的区间，请你计算出在宣传栏上所有贴有海报的连续区间。

编程任务：

根据给定的每张海报所占据的区间，请你计算出在宣传栏上所有贴有海报的连续区间。

数据输入：

由文件input.txt给出输入数据。第一行有1个正整数N，表示海报的总数(3<=N<=50000)。

接下来N行，每行有两个整数left和right，表示海报所占据的区(0<=left<right<=1000000)。

结果输出:

将计算出的区间输出到文件output.txt。每行输出一个连续区间的left与right点，中

间用一空格隔开。请按left点从小到大输出连续区间。

输入示例输出示例

input.txt output.txt

3

1 5

2 8

9 20

1 8

9 20-Column clean-up the issue of publicity

Problem descr iption:

School students in an apartment building, each floor, the lobby bar of publicity every day a large number of posters.

In order to use a special method of cleaning up the boards without leaving traces of ugly, need to know in advance promotional column posted

Posters of the location information.

To simplify the problem we require:

1. Publicity column with a range [0, 1000000] expressed.

2. The assumption that the height of posters and publicity are the same height field, the posters and left with two integer

right (right> left) to indicate that the occupied range of posters [left, right].

3. Two posters can be partially or entirely overlap.

4. All the posters would not be longer than the size of boards.

Each based on a given interval occupied by a poster, please calculate all the columns in the publicity posters put up consecutive interval.

Programming tasks:

Each based on a given interval occupied by a poster,