poj3190

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Stall Reservations

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice POJ 3190

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：
       需要安排N头牛在若干机器上进行工作。每一头牛有它自己的一段工作时间（起始时间s~结束时间t）且每头牛都只能单独在一台机器上进行工作。问至少需要多少台机器。

输入：

       第一行N，之后N行，每行两个整数表示每一头牛工作时间段的起始时刻与结束时刻。

输出：

       最小的机器台数。

分析：

       根据贪心的思想。我们先将牛的时间区间按起始时间进行排序，起始时间小的区间排在前面，起始时间相同的则结束时间小的排在前面。然后定义一个优先队列q，将正在工作的牛对应的区间放入该队列，结束时间小的排在前面。一开始我们遍历每一个牛区间，这时如果我们查找发现优先队列q的队首元素的右区间小于遍历到的牛区间的左端点，则说明之前被一头牛占用的一台机器已经空出来了。那么这时候安排当前牛到那空出来的机器上去工作，否则新增一台机器。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <queue>
 5 using namespace std;
 6 #define MAX_N 50000
 7 int n,use[MAX_N + 1];
 8 struct Node{
 9     int l,r,pos;
10     bool operator <(const Node &a)const {
11         //if(r == a.r) return l > a.l;
12         return r > a.r;
13     }
14 }a[MAX_N + 1];
15 priority_queue<Node> q;
16 bool cmp(Node a,Node b){
17     if(a.l == b.l) return a.r < b.r;
18     return a.l < b.l;
19 }
20 int main(){
21     while(scanf("%d",&n) != EOF){
22         for(int i = 0 ; i < n ; i++){
23             scanf("%d%d",&a[i].l,&a[i].r);
24             a[i].pos = i;
25         }
26         sort(a,a + n,cmp);
27         q.push(a[0]);
28         int now = 0,ans = 1;
29         use[a[0].pos] = 1;
30         for(int i = 1 ; i < n ; i++){
31             if(!q.empty() && q.top().r < a[i].l){
32                 use[a[i].pos] = use[q.top().pos];
33                 q.pop();
34             }
35             else{
36                 ans++;
37                 use[a[i].pos] = ans;
38             }
39             q.push(a[i]);
40         }
41         printf("%d\n",ans);
42         for(int i = 0 ; i < n ; i++)
43             printf("%d\n",use[i]);
44         while(!q.empty()) q.pop();
45     }
46     return 0;
47 }

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poj3190

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原文地址：http://www.cnblogs.com/cyb123456/p/5785882.html