贪心POJ3190:Stall Reservations

Posted 2020-10-24 听风不成泣

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.

 

大致题意：

有一群牛，每个牛都有固定的吃饭时间，并且一个牛槽只能同时让一只牛吃饭。给出所有牛吃饭的开始时间和结束时间，求出最优的策略，使用最少的牛槽让所有的牛吃上饭。输出最少的牛槽数，以及每个牛在第几个牛槽吃。

 

大体思路：

贪心。用到了优先队列。

1)将牛按照开始时间由小到大排序，开始时间相等的话按结束时间由小到大排序。

2)创建一个优先队列，在队列中牛按照结束的时间从小到大排序。

3)把第一只牛加入队列，然后检查第二只牛能否和它同用一槽。如果能，第一只牛出队列，第二只牛进队列。若不能，第二只牛进队列，牛槽数加一。继续该步骤，知道所有的牛都吃过了。

4）注意，记录牛的数据时，除了开始和结束时间，还要记录牛一开始的顺序，以便记录它在哪个槽吃。

 

代码如下：

 1 #include <iostream>
 2 #include <queue>
 3 #include <algorithm>
 4 #include <cstdio>
 5 using namespace std;
 6 const int maxn = 50010;
 7 struct node
 8 {
 9     int st,en,pos;
10     friend const bool operator < (const node a ,const node b)
11     {
12         if(a.en==b.en)
13             return a.st>b.st;
14         return a.en>b.en;
15     }
16 };
17 node cow[maxn];
18 bool cmp(node a,node b)
19 {
20     if(a.st==b.st)
21         return a.en<b.en;
22     return a.st<b.st;
23 }
24 priority_queue<node> q;
25 int ans[maxn];
26 int n;
27 int main()
28 {
29     while(~scanf("%d",&n))
30     {
31         for(int i=1;i<=n;i++)
32         {
33             scanf("%d%d",&cow[i].st,&cow[i].en);
34             cow[i].pos=i;
35         }
36         sort(cow+1,cow+1+n,cmp);
37         q.push(cow[1]);
38         int cnt=1;
39         ans[cow[1].pos]=cnt;
40         for(int i=2;i<=n;i++)
41         {
42             if(q.top().en<cow[i].st)
43             {
44                 ans[cow[i].pos]=ans[q.top().pos];
45                 q.pop();
46             }
47             else
48             {
49                 cnt++;
50                 ans[cow[i].pos]=cnt;
51             }
52             q.push(cow[i]);
53         }
54         printf("%d\n",cnt);
55         for(int i=1;i<=n;i++)
56         {
57             printf("%d\n",ans[i]);
58         }
59     }
60     return 0;
61 }

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