畜栏保留问题（贪心）

畜栏保留问题

Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A…B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows.

Help FJ by determining:
The minimum number of stalls required in the barn so that each cow can have her private milking period
An assignment of cows to these stalls over time
Many answers are correct for each test dataset; a program will grade your answer.

Input

Line 1: A single integer, N

Lines 2…N+1: Line i+1 describes cow i’s milking interval with two space-separated integers.

Output

Line 1: The minimum number of stalls the barn must have.

Lines 2…N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

Sample Input

5
1 10
2 4
3 6
5 8
4 7

Sample Output

4
1
2
3
2
4

Hint

Explanation of the sample:

Here’s a graphical schedule for this output:

Time 1 2 3 4 5 6 7 8 9 10

Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>

Stall 2 … c2>>>>>> c4>>>>>>>>> … …

Stall 3 … … c3>>>>>>>>> … … … …

Stall 4 … … … c5>>>>>>>>> … … …
Other outputs using the same number of stalls are possible.

做的使用超时，暂时直到使用优先队列保证结束时间的有序，可以节省时间

上代码

#include <bits/stdc++.h>

using namespace std;

struct node{
    int no;
    int b;
    int e;
    int a;
    bool operator < (const node &aa)const
    {
        return e>aa.e;
    }
}cattle[50010];
int a[50010];
priority_queue<node> q;

int cmp(node a,node b)
{
    if(a.b==b.b)
        return a.e < b.e;
    return a.b < b.b;
}
int main()
{
    int n,num;
    scanf("%d",&n);
    num = n;
    for(int i = 0;i < n;i++)
    {
        scanf("%d%d",&cattle[i].b,&cattle[i].e);
        cattle[i].no = i;
    }
    sort(cattle,cattle+n,cmp);
    int ans = 0;
    for(int i = 0;i < n;i++)
    {
        if(!q.empty()&&q.top().e < cattle[i].b)
        {
            a[cattle[i].no] = a[q.top().no];
            q.pop();
            q.push(cattle[i]);
        }
        else
        {
            ans++;
            a[cattle[i].no] = ans;
            q.push(cattle[i]);
        }
    }
    printf("%d\n",ans);
    for(int i = 0;i < n;i++)
    {
        printf("%d\n",a[i]);
    }
    return 0;
}