1651: [Usaco2006 Feb]Stall Reservations 专用牛棚
Time Limit: 10 Sec Memory Limit: 64 MB
Submit: 638 Solved: 359
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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
其实就是算每一个时间点有多少只牛在喝水。。求最大的那个就是需要几个水龙头了，上数学课突然就想出来了。。。
所以就容易了。
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<deque>
using namespace std;
int a[1000005];
int main(){
int n;
int nmax=-1;
scanf("%d",&n);
for(int i=1,o,e;i<=n;i++){
scanf("%d%d",&o,&e);
a[o]++;a[e+1]--;
nmax=max(nmax,e);
}
int tmp=0;int ans=-1;
for(int i=1;i<=nmax;i++){
tmp+=a[i];
if(tmp>ans)
ans=tmp;
}
printf("%d\n",ans);
return 0;
}
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