题目
1651: [Usaco2006 Feb]Stall Reservations 专用牛棚
Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 553 Solved: 307
[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
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
题解
这道题用时间戳的思路就可以了，我们统计同一时间最大的时间戳个数就是答案。
代码
1 /*Author:WNJXYK*/ 2 #include<cstdio> 3 using namespace std; 4 5 #define LL long long 6 #define Inf 2147483647 7 #define InfL 10000000000LL 8 9 inline int abs(int x){if (x<0) return -x;return x;} 10 inline int abs(LL x){if (x<0) return -x;return x;} 11 inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;} 12 inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;} 13 inline int remin(int a,int b){if (a<b) return a;return b;} 14 inline int remax(int a,int b){if (a>b) return a;return b;} 15 inline LL remin(LL a,LL b){if (a<b) return a;return b;} 16 inline LL remax(LL a,LL b){if (a>b) return a;return b;} 17 inline void read(int &x){x=0;int f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}x=x*f;} 18 inline void read(LL &x){x=0;LL f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}x=x*f;} 19 inline void read(int &x,int &y){read(x);read(y);} 20 inline void read(LL &x,LL &y){read(x);read(y);} 21 inline void read(int &x,int &y,int &z){read(x,y);read(z);} 22 inline void read(int &x,int &y,int &n,int &m){read(x,y);read(n,m);} 23 inline void read(LL &x,LL &y,LL &z){read(x,y);read(z);} 24 inline void read(LL &x,LL &y,LL &n,LL &m){read(x,y);read(n,m);} 25 const int Maxn=1000000; 26 int n; 27 int a,b; 28 int t[Maxn+10]; 29 int main(){ 30 read(n); 31 for (;n--;){ 32 read(a,b); 33 t[a]++; 34 t[b+1]--; 35 } 36 int Ans=0; 37 for (int i=1;i<=Maxn;i++){ 38 t[i]=t[i-1]+t[i]; 39 Ans=remax(Ans,t[i]); 40 } 41 printf("%d ",Ans); 42 return 0; 43 }