<cf>Schedule

Schedule

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

At the beginning of the new semester there is new schedule in the Berland State University. According to this schedule, n groups have lessons at the room 31. For each group the starting time of the lesson and the finishing time of the lesson are known. It has turned out that it is impossible to hold all lessons, because for some groups periods of their lessons intersect. If at some moment of time one groups finishes it's lesson, and the other group starts the lesson, their lessons don't intersect.

The dean wants to cancel the lesson in one group so that no two time periods of lessons of the remaining groups intersect. You are to find all ways to do that.

Input

The first line contains integer n (1 ≤ n ≤ 5000) — amount of groups, which have lessons in the room 31. Then n lines follow, each of them contains two integers l_i r_i (1 ≤ l_i < r_i ≤ 10⁶) — starting and finishing times of lesson of the i-th group. It is possible that initially no two lessons intersect (see sample 1).

Output

Output integer k — amount of ways to cancel the lesson in exactly one group so that no two time periods of lessons of the remaining groups intersect. In the second line output k numbers — indexes of groups, where it is possible to cancel the lesson. Groups are numbered starting from 1 in the order that they were given in the input. Output the numbers in increasing order.

Sample test(s)

input

3
3 10
20 30
1 3

output

3
1 2 3 

input

4
3 10
20 30
1 3
1 39

output

1
4 

input

3
1 5
2 6
3 7

output

0

AC Code #1:

//Memory: 1500 KB		Time: 50 MS
//Language: GNU C++ 4.6		Result: Accepted
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

struct Group
{
    int s,e,idx;//"s" for start,"e" for end,"idx" for index 
}a[5001];

int pos[5001],k=0;//pos 记录被舍弃的课程的下标
int n,i,j;
bool OK=1;

bool cmp(Group x, Group y)
{
    if(x.e!=y.e) return x.e<y.e;//按结束时间从小到大排序
    else if(x.s!=y.s) return x.s<y.s;
    else return x.idx<y.idx;
}

//此函数用于判断去掉a[i]后余下课程是否有时间冲突，是则返回false  
bool check(int i)
{
    for(int j=1;j<i-1;j++)
    {
        if(a[j].e>a[j+1].s)
        {
            return 0;
        }
    }
    if(i>1 && i<n && a[i-1].e>a[i+1].s)
    {
        return 0;
    }
    for(int j=i+1;j<n;j++)
    {
        if(a[j].e>a[j+1].s)
        {
            return 0;
        }
    }
    return 1;
}

int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i].s,&a[i].e);
        a[i].idx=i;
    }
    sort(a+1,a+n+1,cmp);
    for(i=1;i<n;i++)
    {
        if(a[i].e>a[i+1].s)
        {
            if(!k)
            {
                bool m=check(i);
                bool n=check(i+1);
                if(m && !n) pos[k++]=a[i].idx;
                else if(!m && n)
                {
                    pos[k++]=a[i+1].idx;
                    i++;
                }
                else if(m && n)
                {
                    pos[k++]=a[i].idx;
                    pos[k++]=a[i+1].idx;
                    i++;
                }
                else
                {
                    OK=0;
                    break;
                }
            }
            else
            {
                OK=0;
                break;
            }
        }
    }
    if(!OK) puts("0");
    //ok==1但k==0
    else if(!k)
    {
        printf("%d\n",n);
        for(i=1;i<n;i++)
            printf("%d ",i);
        printf("%d\n",i);
    }
    //ok==1 and k>0
    else
    {
        sort(pos,pos+k);
        printf("%d\n",k);
        for(i=0;i<k-1;i++)
            printf("%d ",pos[i]);
        printf("%d\n",pos[i]);
    }
    return 0;
}

还有一个复杂一点的方法：

AC Code #2：

//Memory: 1500 KB		Time: 50 MS
//Language: GNU C++ 4.6		Result: Accepted

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

struct Group
{
    int s,e,idx;
}a[5001];

int pos[5001],k=0;
int n,i,j;
bool OK=1;

bool cmp(Group x, Group y)
{
    if(x.e!=y.e) return x.e<y.e;//按结束时间从小到大排序
    else if(x.s!=y.s) return x.s<y.s;
    else return x.idx<y.idx;
}

int check(int i)
{
    int j=i+1;
    while(a[i].e>a[j].s && j<=n)
    {
        j++;
    }
    return j;
}

int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i].s,&a[i].e);
        a[i].idx=i;
    }
    sort(a+1,a+n+1,cmp);
    for(i=1;i<n;i++)
    {
        j=check(i);
        //a[i]的结束时间晚于a[i+1]和a[i+2]的开始时间，必弃a[i]
        if(j-i>2)
        {
            //当前面已有课程被舍去，OK为假.
            if(k)
            {
                OK=0;
                break;
            }
            else
            {
                pos[k]=a[i].idx;
                //当后面仍有课程必须舍弃，OK也为假
                for(i++;i<n && a[i].e<=a[i+1].s;i++){}
                if(i<n)
                {
                    OK=0;
                }
                break;
            }
        }
        //这时a[i]的结束时间大于且仅大于a[i+1]的开始时间,两者要舍去其一
        else if(j-i==2)
        {
            //当前面已有课程被舍去，OK为假.
            if(k)
            {
                OK=0;
                break;
            }
            else
            {
                //当>2时a[i+1]必定要舍去;当==2时，a[i+1]和a[i+2]必弃其一，又结合
                //a[i]和a[i+1]必弃其一，则还是只能舍弃a[i+1].
                if(check(i+1)-(i+1)>=2)
                {
                    pos[k++]=a[i+1].idx;
                }
                else if(i>0 && a[i-1].e>a[i+1].s)
                {
                    pos[k++]=a[i+1].idx;
                }
                //此时a[i+1].e<=a[i+2].s,a[i],a[i+1]都可以选为被舍弃的组。
                else
                {
                    pos[k++]=a[i].idx;
                    pos[k++]=a[i+1].idx;
                }
                i++;
            }
        }
    }
    if(!OK) puts("0");
    //ok==1但k==0
    else if(!k)
    {
        printf("%d\n",n);
        for(i=1;i<n;i++)
            printf("%d ",i);
        printf("%d\n",i);
    }
    else
    {
        sort(pos,pos+k);
        printf("%d\n",k);
        for(i=0;i<k-1;i++)
            printf("%d ",pos[i]);
        printf("%d\n",pos[i]);
    }
    return 0;
}