贪心：zoj3953 Intervals

Description

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.

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

Sample Output

4
3 5 7 10

题意：

　　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。

　　如果不相交则删除最左侧的区间然后继续添加区间，每三个判断一次。

代码：

#include <bits/stdc++.h>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>

#define IO ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
const long long INF = 0x3f3f3f3f;
const long long mod = 1e9+7;
const double PI = acos(-1.0);
const double wyth=(sqrt(5)+1)/2.0;
const char week[7][10]= {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};
const char month[12][10]= {"Janurary","February","March","April","May","June","July",
                           "August","September","October","November","December"
                          };
const int daym[2][13] = {{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
    {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
};
const int dir4[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
const int dir8[8][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1}};
using namespace std;
const int maxn = 50005;
int ans[maxn],cnt;
struct node
{
    int l,r;
    int id;
} a[maxn], tmp[5];
bool cmp1(node a, node b)//按起点从小到大排序
{
    if(a.l == b.l)
        return a.r < b.r;
    return a.l < b.l;
}
bool cmp2(node a, node b)//按终点从大到小排列
{
    return a.r > b.r;
}
bool f(node x, node y, node z)//判断是否相交
{
    return y.l <= x.r && z.l <= y.r && z.l <= x.r;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cnt = 0;
        int n;
        cin>>n;
        for(int i = 1; i <= n; i++)
        {
            cin>>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);
            //如果两两相交，记录下来，然后将最右侧的区间交换到tmp[3]
            if(f(tmp[1], tmp[2], tmp[3]))
            {
                sort(tmp+1, tmp+4, cmp2);
                ans[cnt++] = tmp[1].id;
                swap(tmp[1], tmp[3]);
            }
            //如果不相交，那么将最左侧的区间交换到tmp[3];
            else
                sort(tmp+1, tmp+4, cmp2);
        }
        sort(ans, ans+cnt);
        cout<<cnt<<endl;
        for(int i = 0; i < cnt; i++)
            cout<<ans[i]<<" ";
        cout<<endl;
    }
    return 0;
}