ZOJ - 3953Intervals(贪心)

Intervals
Time Limit: 1 Second      Memory Limit: 65536 KB      Special Judge
Chiaki has n intervals and the i-th of them is [li, ri].
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 la ≤ x ≤ ra and lb ≤ x ≤ rb.

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 li and ri (1 ≤ li < ri ≤
109) denoting the i-th interval. Note that for every 1 ≤ i < j ≤ n, li ≠ lj or ri ≠ rj.
It is guaranteed that the sum of all n does not exceed 500000.

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.

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

Sample Output

4

3 5 7 10题目大意：
给你n个区间的左右端点，让你取走一些区间，使得任何三个区间都不两两相交。
思路：
贪心，首先将所有区间以左端点从小到大排序，然后每次判断当前的三个区间是否两两相交，如果两两相交则删除右端点值最大的那个区间，因为这样对后面的影响会最小。如果不相交则继续添加区间，每三个一判断。#include<bits/stdc++.h>
using namespace std;
const int maxn = 50005;
struct Intervals
{
int l, r, id;
}a[maxn], tmp[5];
bool cmp1(Intervals p, Intervals q)
{
if(p.l == q.l)
return p.r < q.r;
return p.l < q.l;
}
bool cmp2(Intervals p, Intervals q)
{
return p.r > q.r;
}
bool isInterval(Intervals x, Intervals y, Intervals z)
{
return y.l <= x.r && z.l <= y.r && z.l <= x.r;
}
int ans[maxn], cnt;
int main()
{
int n, T;
scanf("%d", &T);
while(T--)
{
cnt = 0;
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%d%d", &a[i].l, &a[i].r);
a[i].id = i;
}
sort(a+1, a+n+1, cmp1);
tmp[1] = a[1];
tmp[2] = a[2];
for(int i = 3; i <= n; i++)
{
tmp[3] = a[i];
sort(tmp+1, tmp+4, cmp1);
if(isInterval(tmp[1], tmp[2], tmp[3]))
{
sort(tmp+1, tmp+4, cmp2);
ans[cnt++] = tmp[1].id;
swap(tmp[1], tmp[3]);
}
else
sort(tmp+1, tmp+4, cmp2);
}
sort(ans, ans+cnt);
printf("%d\n", cnt);
for(int i = 0; i < cnt; i++)
printf("%d%c", ans[i], i == cnt-1 ? '\n' : ' ');
}
return 0;
}