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

Posted 2020-10-06 Lev今天学习了吗

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

有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall才能满足它们的要求

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


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.

HINT

 

不妨试下这个数据,对于按结束点SORT,再GREEDY的做法 1 3 5 7 6 9 10 11 8 12 4 13 正确的输出应该是3

 

Source

Silver

 

Analysis

求区间叠加最大数，经典的差分老题

然而第一次提交的时候因为是扫描 1-n 不知道为什么被卡了

后来改成了 1-最远区间右端点 就过了

很迷

 

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