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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
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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

题解:
每个时间点被覆盖的次数取max的就是答案。
原来差分序列可以这么用。长见识了。
将原序列差分了之后,前缀和代表改点的值,既不用打线段树了。
代码:
 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 100000000013 #define maxn 100001014 #define maxm 500+10015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 using namespace std;19 inline int read()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);31     n=read();32     for(int i=1;i<=n;i++)33     {34      int x=read(),y=read()+1;    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 }
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BZOJ1651: [Usaco2006 Feb]Stall Reservations 专用牛棚