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:
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
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.
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.
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 10 Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>> Stall 2 .. c2>>>>>> c4>>>>>>>>> .. .. Stall 3 .. .. c3>>>>>>>>> .. .. .. .. Stall 4 .. .. .. c5>>>>>>>>> .. .. ..Other outputs using the same number of stalls are possible.
大致题意：
有一群牛，每个牛都有固定的吃饭时间，并且一个牛槽只能同时让一只牛吃饭。给出所有牛吃饭的开始时间和结束时间，求出最优的策略，使用最少的牛槽让所有的牛吃上饭。输出最少的牛槽数，以及每个牛在第几个牛槽吃。
大体思路：
贪心。用到了优先队列。
1)将牛按照开始时间由小到大排序，开始时间相等的话按结束时间由小到大排序。
2)创建一个优先队列，在队列中牛按照结束的时间从小到大排序。
3)把第一只牛加入队列，然后检查第二只牛能否和它同用一槽。如果能，第一只牛出队列，第二只牛进队列。若不能，第二只牛进队列，牛槽数加一。继续该步骤，知道所有的牛都吃过了。
4）注意，记录牛的数据时，除了开始和结束时间，还要记录牛一开始的顺序，以便记录它在哪个槽吃。
代码如下：
1 #include <iostream> 2 #include <queue> 3 #include <algorithm> 4 #include <cstdio> 5 using namespace std; 6 const int maxn = 50010; 7 struct node 8 { 9 int st,en,pos; 10 friend const bool operator < (const node a ,const node b) 11 { 12 if(a.en==b.en) 13 return a.st>b.st; 14 return a.en>b.en; 15 } 16 }; 17 node cow[maxn]; 18 bool cmp(node a,node b) 19 { 20 if(a.st==b.st) 21 return a.en<b.en; 22 return a.st<b.st; 23 } 24 priority_queue<node> q; 25 int ans[maxn]; 26 int n; 27 int main() 28 { 29 while(~scanf("%d",&n)) 30 { 31 for(int i=1;i<=n;i++) 32 { 33 scanf("%d%d",&cow[i].st,&cow[i].en); 34 cow[i].pos=i; 35 } 36 sort(cow+1,cow+1+n,cmp); 37 q.push(cow[1]); 38 int cnt=1; 39 ans[cow[1].pos]=cnt; 40 for(int i=2;i<=n;i++) 41 { 42 if(q.top().en<cow[i].st) 43 { 44 ans[cow[i].pos]=ans[q.top().pos]; 45 q.pop(); 46 } 47 else 48 { 49 cnt++; 50 ans[cow[i].pos]=cnt; 51 } 52 q.push(cow[i]); 53 } 54 printf("%d\n",cnt); 55 for(int i=1;i<=n;i++) 56 { 57 printf("%d\n",ans[i]); 58 } 59 } 60 return 0; 61 }
