Stall Reservations（贪心，****思路理顺）

最新推荐文章于 2022-03-12 12:54:17 发布

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最新推荐文章于 2022-03-12 12:54:17 发布

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本文链接：https://blog.youkuaiyun.com/qq_37280023/article/details/75414755

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本文介绍了一种解决奶牛饮水隔间分配问题的有效算法。通过将奶牛按照开始饮水时间排序，并采用优先队列来跟踪隔间的使用情况，实现了最小隔间数量的分配方案。文章还提供了实现代码。

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Stall Reservations

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.

题意：

有n头奶牛，每头奶牛喝牛奶的时候都不愿意被打扰，告诉你每头奶牛开始喝牛奶和结束和牛奶的时间 a 和 b，求至少需要多少个隔间，使每头奶牛喝牛奶的时间不相交

输入：

第一行 n
接下来n行，第n-1头奶牛的时间 a b

输出：

第一行：最小的隔间数
接下来n行：每头奶牛对应的隔间标号（只输出一种情况即可）

思路：

每头奶牛按照开始的时间从小到大排，每一次如果能安排进最早结束的隔间，则刷新此隔间的结束时间，否则开一个新的隔间
奶牛用结构体存储开始，结束时间和隔间号

注意

代码的结构和时间要求

#include <iostream>
#include <cstring>
#include <map>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
struct cow
{
    int s,e,num,id;
    cow(int s=0,int e=0,int num=0):s(s),e(e),num(num) {};
};
bool cmp (const cow& a,const cow& b){
    return a.num<b.num;
}
bool operator < (const cow& a,const cow &b){
    return a.s<b.s;
}
cow c[55000];
typedef pair<int,int> Pir;

int main()
{
    //freopen("1.txt","r",stdin);
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++){
            int a,b; scanf("%d%d",&a,&b);
            c[i]=cow(a,b,i+1);    //初始化n头奶牛
        }
        sort(c,c+n);
        int room=0;
        priority_queue<Pir,vector<Pir>,greater<Pir> > que; //每次取结束时间，即.first最小的隔间
        que.push(Pir(c[0].e,++room));
        c[0].id=room;
        for(int i=1;i<n;i++)
        {
                Pir p=que.top();
                if(p.first<c[i].s){ //如果此时开始时间最早的奶牛能安排进结束时间最早的隔间
                    que.pop();
                    que.push(Pir(c[i].e,p.second));
                    c[i].id=p.second; //记录排进的隔间编号
                }   else{           //如果不能
                    que.push(Pir(c[i].e,++room));
                    c[i].id=room;
                }
        }

        cout<<room<<endl;
        sort(c,c+n,cmp);   //按输入顺序重新排序，便于输出
        for(int i=0;i<n;i++)
            printf("%d\n",c[i].id);
    }
    return 0;
}