POJ 3190 Stall Reservations

qldabiaoge 2018-03-22 原文

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.

一群奶牛，只在固定的时间产奶，每头牛只在固定的时间产奶，每头牛都需要一个挤奶的机器

所有奶牛产奶，最少需要多少个机器，输出每一个奶牛使用挤奶机器的编号

这题贪心，需要用优先队列优化，

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn=50010;
 7 struct node {
 8     int x,y,id;
 9     friend bool operator<(node a,node b) {
10         return a.y>b.y;
11     }
12 } qu[maxn];
13 int cmp(node a,node b) {
14     return a.x<b.x;
15 }
16 int ID[maxn];
17 int main() {
18     int n;
19     while(scanf("%d",&n)!=EOF ) {
20         for (int i=0 ; i<n ; i++) {
21             scanf("%d%d",&qu[i].x,&qu[i].y);
22             qu[i].id=i;
23         }
24         sort(qu,qu+n,cmp);
25         memset(ID,0,sizeof(ID));
26         priority_queue<node>q;
27         int ans=1;
28         q.push(qu[0]);
29         ID[qu[0].id]=1;
30         for (int i=1 ; i<n ; i++) {
31             if (q.top().y<qu[i].x) {
32                 ID[qu[i].id]=ID[q.top().id];
33                 q.pop();
34                 q.push(qu[i]);
35             } else {
36                 ID[qu[i].id]=++ans;
37                 q.push(qu[i]);
38             }
39         }
40         printf("%d\n",ans);
41         for (int i=0 ; i<n ; i++) {
42             printf("%d\n",ID[i]);
43         }
44     }
45     return 0;
46 }