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poj 3190 Stall Reservations(贪心)

时间:2014-07-06 14:53:48      阅读:130      评论:0      收藏:0      [点我收藏+]

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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.
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",n);
55     for(i=1; i<=a; i++)
56         printf("%d\n",s[i]);
57 }

 

 有什么问题可以评论提问。

poj 3190 Stall Reservations(贪心),布布扣,bubuko.com

poj 3190 Stall Reservations(贪心)

标签:des   style   blog   color   os   问题   

原文地址:http://www.cnblogs.com/Xacm/p/3825593.html

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