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 Many answers are correct for each test dataset; a program will grade your answer.
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 1 2 3 2 4
Hint
Explanation of the sample:
Here's a graphical schedule for this output:
Here's a graphical schedule for this output:
Time 1 2 3 4 5 6 7 8 9 10Other outputs using the same number of stalls are possible.
Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..
Stall 3 .. .. c3>>>>>>>>> .. .. .. ..
Stall 4 .. .. .. c5>>>>>>>>> .. .. ..
Source
一些奶牛要在指定的时间内挤牛奶，而一个机器只能同时对一个奶牛工作。给你每头奶牛的指定时间的区间，问你最小需要多少机器。
先按起始时间从小到大排序，起始时间相同则按结束时间从小到大排序
然后用一个优先队列来维护，vis记录奶牛的工作机器
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<set> 5 #include<map> 6 #include<algorithm> 7 #include<queue> 8 using namespace std; 9 #define N 60000 10 int n; 11 int vis[N]; 12 struct Node 13 { 14 int l,r; 15 int pos; 16 bool friend operator < (Node a,Node b) 17 { 18 if(a.r!=b.r) 19 return a.r>b.r; 20 } 21 }p[N]; 22 bool cmp(Node a,Node b) 23 { 24 if(a.l!=b.l) 25 return a.l<b.l; 26 return a.r<b.r; 27 } 28 int main() 29 { 30 while(scanf("%d",&n)==1) 31 { 32 memset(vis,0,sizeof(vis)); 33 for(int i=0;i<n;i++) 34 { 35 scanf("%d%d",&p[i].l,&p[i].r); 36 p[i].pos=i; 37 } 38 sort(p,p+n,cmp); 39 40 int ans=1; 41 priority_queue<Node>q; 42 q.push(p[0]); 43 vis[p[0].pos]=ans; 44 45 for(int i=1;i<n;i++) 46 { 47 Node tmp=q.top(); 48 if(tmp.r<p[i].l){ 49 q.pop(); 50 vis[p[i].pos]=vis[tmp.pos]; 51 } 52 else{ 53 ans++; 54 vis[p[i].pos]=ans; 55 } 56 q.push(p[i]); 57 } 58 printf("%d ",ans); 59 for(int i=0;i<n;i++) printf("%d ",vis[i]); 60 61 } 62 return 0; 63 }