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();\

    cout.tie();

const long long INF = 0x3f3f3f3f;

const long long mod = 1e9+;

const double PI = acos(-1.0);

const double wyth=(sqrt()+)/2.0;

const char week[][]= {"Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"};

const char month[][]= {"Janurary","February","March","April","May","June","July",

                           "August","September","October","November","December"

                          };

const int daym[][] = {{, , , , , , , , , , , , },

    {, , , , , , , , , , , , }

};

const int dir4[][] = {{, }, {, }, {-, }, {, -}};

const int dir8[][] = {{, }, {, }, {-, }, {, -}, {, }, {-, -}, {, -}, {-, }};

using namespace std;

const int maxn = ;

int ans[maxn],cnt;

struct node

{

    int l,r;

    int id;

} a[maxn], tmp[];

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 = ;

        int n;

        cin>>n;

        for(int i = ; i <= n; i++)

        {

            cin>>a[i].l>>a[i].r;

            a[i].id = i;

        }

        sort(a+, a+n+, cmp1);

        tmp[] = a[];

        tmp[] = a[];

        for(int i = ; i <= n; i++)

        {

            tmp[] = a[i];

            sort(tmp+, tmp+, 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[], tmp[], tmp[]))

            {

                sort(tmp+, tmp+, cmp2);

                ans[cnt++] = tmp[].id;

                swap(tmp[], tmp[]);

            }

            //如果不相交，那么将最左侧的区间交换到tmp[3];

            else

                sort(tmp+, tmp+, cmp2);

        }

        sort(ans, ans+cnt);

        cout<<cnt<<endl;

        for(int i = ; i < cnt; i++)

            cout<<ans[i]<<" ";

        cout<<endl;

    }

    return ;

}