# BZOJ1651: [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: 509  Solved: 280
[Submit][Status]

## 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.

## Source

 1 #include<cstdio>
2 #include<cstdlib>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 #include<iostream>
7 #include<vector>
8 #include<map>
9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 1000010
14 #define maxm 500+100
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 using namespace std;
20 {
21     int x=0,f=1;char ch=getchar();
22     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
23     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
24     return x*f;
25 }
26 int n,sum=0,ans=0,mx=0,a[maxn];
27 int main()
28 {
29     freopen("input.txt","r",stdin);
30     freopen("output.txt","w",stdout);
32     for(int i=1;i<=n;i++)
33     {
35      a[x]++,a[y]--;
36      if(y>mx)mx=y;
37     }
38     for(int i=1;i<=mx;i++)
39      {
40          sum+=a[i];
41          if(sum>ans)ans=sum;
42      }
43     printf("%d\n",ans);
44     return 0;
45 }
View Code

posted @ 2014-08-29 15:16  ZYF-ZYF  Views(170)  Comments(0Edit  收藏  举报