# [BZOJ] 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

## 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 987  Solved: 566
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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

## 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.

5
1 10
2 4
3 6
5 8
4 7

## Sample Output

4

OUTPUT DETAILS:

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.

Silver

## Code

 1 #include<cstdio>
2 #include<iostream>
3 using namespace std;
4
5 int r,cnt[10000000],ret,cncct,a,b,n;
6
7 int main(){
8     scanf("%d",&n);
9
10     for(int i = 1;i <= n;i++){
11         scanf("%d %d",&a,&b);
12         cnt[a]++,cnt[b+1]--;
13         r = max(r,b);
14     }
15
16     int cncct = 0,ret = 0;
17     for(int i = 1;i <= r;i++){
18         cncct += cnt[i];
19         ret = max(ret,cncct);
20     }
21
22     printf("%d",ret);
23
24     return 0;
25 }
84 ms 的差分

posted @ 2017-09-10 11:27  Leviaton  阅读(169)  评论(0编辑  收藏  举报