3953. Intervals

Intervals

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

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题意

给定若干个区间(保证左端点都不相同，且右端点都不相同)，判断最少需要删除几个区间，才能使得剩下的区间中任取三个都不会两两相交，并列出删除的区间的编号。

思路

区间贪心问题。先将所有区间按左端点从小到大排列，再依次选三个区间判断是否两两相交：①若两两相交，则将这三个区间中右端点最大的区间删除(这样可以最大程度的减少区间影响范围)，加入下一个区间再进行比较；②若不满足两两相交，则用下一个区间替换三个区间中右端点最小的区间(必然不可能与下个区间相交)再进行比较。
最后需要注意输出格式的处理。

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代码实现

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 50001;

// 定义区间
struct node
{
    int x, y;
    int id;
} interval[maxn];

// 按照左端点从小到大排列
bool cmpL(node a, node b)
{
    return a.x < b.x;
}

// 按照右端点从小到大排列
bool cmpR1(node a, node b)
{
    return a.y < b.y;
}

// 按照右端点从大到小排列
bool cmpR2(node a, node b)
{
    return a.y > b.y;
}

// 判断是否两两相交
bool judge(node a, node b, node c)
{
    return b.x <= a.y && c.x <= a.y && c.x <= b.y;
}

int main()
{
    int t, n;
    vector<int> ans;        // 删除区间编号记录
    node tmp[4];            // 比较区间
    scanf("%d", &t);
    for (int i = 0; i < t; i++)
    {
        scanf("%d", &n);
        int x, y;
        for (int j = 1; j <= n; j++)
        {
            scanf("%d %d", &x, &y);
            interval[j].x = x;
            interval[j].y = y;
            interval[j].id = j;
        }

        ans.clear();
        sort(interval + 1, interval + n + 1, cmpL);     // 先按左端点排序
        tmp[1] = interval[1];
        tmp[2] = interval[2];
        for (int k = 3; k <= n; k++)
        {
            tmp[3] = interval[k];
            sort(tmp + 1, tmp + 4, cmpL);   // 排序便于判断相交
            // 两种相交情况处理
            if (judge(tmp[1], tmp[2], tmp[3]))
            {
                sort(tmp + 1, tmp + 4, cmpR1);
                ans.push_back(tmp[3].id);
            }
            else
                sort(tmp + 1, tmp + 4, cmpR2);
        }

        // 输出处理
        sort(ans.begin(), ans.end());
        printf("%d\n", ans.size());
        if (ans.size() == 0)
            printf("\n");
        else
        {
            for (int j = 0; j < ans.size(); j++)
            {
                if (j > 0)
                    printf(" ");
                printf("%d", ans[j]);
            }
            printf("\n");
        }
    }
}

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参考

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