首页 > 代码库 > BZOJ1651: [Usaco2006 Feb]Stall Reservations 专用牛棚
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 MBSubmit: 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
1 10
2 4
3 6
5 8
4 7
Sample Output
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
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 }
BZOJ1651: [Usaco2006 Feb]Stall Reservations 专用牛棚