V - 22

最小覆盖区间元素

最新推荐文章于 2024-11-24 11:53:09 发布

原创最新推荐文章于 2024-11-24 11:53:09 发布 · 240 阅读

0 ·

CC 4.0 BY-SA版权

二分贪心专栏收录该内容

14 篇文章

订阅专栏

本文介绍了一个寻找最少元素数量的问题，这些元素至少包含每个指定区间内的两个不同整数。通过将区间按结束点排序并采用贪心算法，文章提供了一种有效的解决方案。

Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b.
Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.

Input

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an interval.

Output

Output the minimal number of elements in a set containing at least two different integers from each interval.

Sample Input

Sample Output

题意：
找一个数组，使数组里至少有两个整数属于间隔，每个间隔里的两个数都在这个数组

分析：
贪心，先按照b从小到大排序，在将数组里的数跟后面的间隔前两个比较

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
	int x;
	int y;
}b[100001];
bool cmp(node a,node b)
{
	return a.y<b.y;
}
int main()
{
	int n,j,i,m[10001];
	scanf("%d",&n);
	for(i=0;i<n;i++)
	scanf("%d%d",&b[i].x,&b[i].y);
	sort(b,b+n,cmp);
	m[0]=b[0].y-1;
	m[1]=b[0].y;
	j=0;
	for(i=1;i<n;i++)
	{
	  if(b[i].x<=m[j]&&m[j+1]<=b[i].y)
	     {
		continue;
	     }
	   if(m[j+1]>=b[i].x&&m[j]<b[i].x)
	{
	   m[j+2]=b[i].y;
	   j=j+1;
	}
	    if(m[j+1]<b[i].x)
	{
	   
	   m[j+2]=b[i].y-1;
	   m[j+3]=b[i].y;	
	   j=j+2;
	}
	}
	printf("%d\n",j+2);
	
}

感受：
挺简单的，就是题意一开始没有读懂。。。。