poj 3190 Stall Reservations（贪心）

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.

题意：奶牛要在指定的时间挤奶，一台机器只能给一头奶牛挤奶，问最少要几台机器。

思路：用STL里的优先队列，判断空闲机器是否满足下一头奶牛的挤奶时间，满足就让挤完奶的奶牛出队把机器编号给下一头奶牛，下一头奶牛入队，否则机器+1，下一头奶牛入队.

注意：输出的顺序和输入的顺序相同，在输入的时候把下标一起存入结构体，另外开一个数组记录机器的编号，在排序后下标就不会乱了。

ac代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<queue>
 6 using namespace std;
 7 struct node
 8 {
 9     int a;
10     int b;
11     int c;
12     bool operator<(const node &c)const
13     {
14         if(b==c.b)
15             return a>c.a;
16         return b>c.b;
17     }
18 } d[60000];
19 priority_queue<node> q;
20 int s[60000];
21 bool cmp(struct node a,struct node b)
22 {
23     if(a.a==b.a)
24         return a.b<b.b;
25     return a.a<b.a;
26 }
27 int main()
28 {
29     int a,b,i,j,n=1;
30     scanf("%d",&a);
31     for(i=1; i<=a; i++)
32     {
33         scanf("%d%d",&d[i].a,&d[i].b);
34         d[i].c=i;
35     }
36     sort(d+1,d+a+1,cmp);
37     q.push(d[1]);
38     s[d[1].c]=1;
39     for(i=2; i<=a; i++)
40     {
41         int x=q.top().b;
42         if(x<d[i].a)
43         {
44             s[d[i].c]=s[q.top().c];
45             q.pop();
46         }
47         else
48         {
49             n++;
50             s[d[i].c]=n;
51         }
52         q.push(d[i]);
53     }
54     printf("%d
",n);
55     for(i=1; i<=a; i++)
56         printf("%d
",s[i]);
57 }

有什么问题可以评论提问。