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poj 3190 Stall Reservations（贪心）

2024-07-10 21:38:51   223人阅读

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

51 102 43 65 84 7

Sample Output

412324

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)const13     {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         else48         {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 }

 

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