E - Intervals 贪心

Chiaki has n intervals and the i-th of them is [l_i, r_i]. She wants to delete some intervals so that there does not exist three intervals a, b and c such that a intersects with b, b intersects with c and c intersects with a.

Chiaki is interested in the minimum number of intervals which need to be deleted.

Note that interval a intersects with interval b if there exists a real number x such that l_a ≤ x ≤ r_a and l_b ≤ x ≤ r_b.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1 ≤ n ≤ 50000) -- the number of intervals.

Each of the following n lines contains two integers l_i and r_i (1 ≤ l_i < r_i ≤ 10⁹) denoting the i-th interval. Note that for every 1 ≤ i < j ≤ n, l_i ≠ l_j or r_i ≠ r_j.

It is guaranteed that the sum of all n does not exceed 500000.

<h4< dd="">Output

For each test case, output an integer m denoting the minimum number of deletions. Then in the next line, output m integers in increasing order denoting the index of the intervals to be deleted. If m equals to 0, you should output an empty line in the second line.

<h4< dd="">Sample Input

1
11
2 5
4 7
3 9
6 11
1 12
10 15
8 17
13 18
16 20
14 21
19 22

<h4< dd="">Sample Output

4
3 5 7 10

E - Intervals
题目大意：
就是给你n组数据，在这n组数据里，删除一部分，使得任意三个组数据不相交。
题目思路：
先对它开始的位置进行排序，小的在前，然后取三组数据，进行判断是否两两相交，如果是就先要对它进行排序。
这个排序比较重要，是以结束的数据为标准，数据越大排在越前面。
如果两两相交就删去之后排序在最前面的那组数据，否则就更新一个为最后那组数据。

其实比较想清楚就比较简单了，分成两组数据进行处理，第一组以l进行排序，第二组以r进行排序，如果要删除，就删除r最大的
那组数据，再更新一个数据。值得学习的是，这种处理方法，这样子可以很好的处理表示删除的数据。

不过这里要注意一下，对是不是两两相交判断的标准，就是先选两个例如x，y，先满足x与y相交，再加入z，z与x，z与y相交

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=50050;
struct node
{
	int l,r,id;
}ex[maxn];
int ans[maxn];

bool cmp1(node a,node b)
{
	if(a.l==b.l) return a.r<b.r;
	return a.l<b.l;
}
bool cmp2(node a,node b)
{
	if(a.r==b.r) return a.l<b.l;
	return a.r>b.r;
}
int isinterval(node x,node y,node z)
{
	int f1=0,f2=0;
	if(x.r>=y.l) f1=1;//x&y
	if(x.r>=z.l&&y.r>=z.l) f2=1;//z&x,z&y
    if(f1&&f2) return 1;
    return 0;
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,pos=0;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&ex[i].l,&ex[i].r);
			ex[i].id=i+1;
		}
		sort(ex,ex+n,cmp1);
		node x[5];
		x[0]=ex[0];
		x[1]=ex[1];
		for(int i=2;i<n;i++)
		{
			x[2]=ex[i];
		//	sort(x,x+3,cmp1);
			int f=isinterval(x[0],x[1],x[2]);
			sort(x,x+3,cmp2);
			if(f)
			{
				ans[pos++]=x[0].id;
				swap(x[0],x[2]);
			}
		}
		sort(ans,ans+pos);
		printf("%d\n",pos);
		if(pos==0) printf("\n");
		else
		{
			for(int i=0;i<pos-1;i++) printf("%d ",ans[i]);
		    printf("%d\n",ans[pos-1]);
		}
	}
	return 0;
}

原文地址：https://www.cnblogs.com/EchoZQN/p/10372040.html