Intervals

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Problem 60: Intervals

Time Limit:1 Ms| Memory Limit:10 MB
Difficulty:0

[Problem] [Rank]

Description

   There is given the series of n closed intervals [ai; b i], where i=1,2,...,n. The sum of those intervals may berepresented as a sum of closed pairwise non−intersecting intervals. The task is to find such representation with the minimal number of intervals. The intervals of this representation should be written in the output file in acceding order. We say that the intervals [a; b] and [c; d] are in ascending order if, and only if a <= b < c <= d.
  
 Task
  
 Write a program which:
  
 .reads from the std input the description of the series of intervals,
  
 .computes pairwise non−intersecting intervals satisfying the conditions given above, 
  
 .writes the computed intervals in ascending order into std output

Input

   In the first line of input there is one integer n, 3 <= n <= 50000. This is the number of intervals. In the (i+1)−st line, 1 <= i <= n, there is a description of the interval [ai; bi] in the form of two integers ai and bi separated by a single space, which are respectively the beginning and the end of the interval,1 <= ai <= bi <= 1000000.

Output

   The output should contain descriptions of all computed pairwise non−intersecting intervals. In each line should be written a description of one interval. It should be composed of two integers, separated by a single space, the beginning and the end of the interval respectively. The intervals should be written into the output in ascending order.

Sample Input

Sample Output

4
  
10

思路：找最大不重合空间，一道水题，一时没思路，浪费了好长时间。。。，先按每

个区间的start排序，然后不断扩充区间（end），如果start > end则输出，注意最后要

把最后一个输出（没有start去比了）。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

using namespace std;

typedef struct Elem
{
	int start;
	int end;
}elem;

elem temp[50001];

bool cmp(elem a, elem b)
{
	if(a.start != b.start)
	{
		return a.start < b.start;
	}
	else
	{
		return a.end < b.end;
	}
}

int main()
{
	int n, i;
	scanf("%d", &n);
	for(i = 0; i < n; i++)
	{
		scanf("%d%d", &temp[i].start, &temp[i].end);	
	}	
	sort(temp, temp+n, cmp);
	elem t = temp[0];
	for(i = 1; i < n; i++)
	{
		if(temp[i].start > t.end)
		{
			printf("%d %d\n", t.start, t.end);
			t = temp[i];
		}
		else
		{
			if(temp[i].end > t.end)
			{
				t.end = temp[i].end;
			}
		}
	}
	printf("%d %d\n", t.start, t.end);
	return 0;
}