Stall Reservations

最新推荐文章于 2021-11-11 18:31:11 发布

原创最新推荐文章于 2021-11-11 18:31:11 发布 · 691 阅读

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本文介绍了一种解决奶牛在特定时间内挤奶的问题的算法。通过优先队列跟踪挤奶位的使用情况，实现了最少挤奶位的高效利用，并确保每头奶牛都能在指定的时间段内得到独立的挤奶位置。

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题目链接
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

Sample Output

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.

这里写图片描述

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
struct Cow{
    int start,end,No;
    bool friend operator < (Cow a,Cow b){
        return a.start < b.start;
    }
}cow[50005];
struct Stall{
    int No,end;
    bool friend operator < (Stall a,Stall b){
        return a.end > b.end;
    }
    Stall (int e,int n):end(e),No(n){}
};
int main(){
    int n,total,cnt,pos[50005];
    priority_queue <Stall> pq;
    scanf("%d",&n);
    for(int i = 0;i < n;i++){
        scanf("%d%d",&cow[i].start,&cow[i].end);
        cow[i].No = i;//记录奶牛的编号,因为只要一进行排序,奶牛本身的编号就会乱掉
    }
    sort(cow,cow + n);
    total = 0;
    for(int i = 0;i < n;i++){
        if(pq.empty()){
            pos[cow[i].No] = ++total;
            pq.push(Stall(cow[i].end,total));//记录栅栏的编号,同时记录栅栏的使用结束时间
        }
        else{
            Stall st = pq.top();
            if(st.end < cow[i].start){
                pq.pop();
                pos[cow[i].No] = st.No;
                pq.push(Stall(cow[i].end,st.No));
            }
            else{
                pos[cow[i].No] = ++total;
                pq.push(Stall(cow[i].end,total));
            }
        }
    }
    printf("%d\n",total);
    for(int i = 0;i < n;i++)
        printf("%d\n",pos[i]);
    return 0;
}