题目链接
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 7Sample Output
4
1
2
3
2
4Hint
Explanation of the sample:
Here’s a graphical schedule for this output:
Time 1 2 3 4 5 6 7 8 9 10Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>Stall 2 ...c2>>>>>>.c4>>>>>>>>>......Stall 3 ......c3>>>>>>>>>............Stall 4 .........c5>>>>>>>>>.........Other outputs using the same number of stalls are possible.
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
struct Cow{int start,end,No;bool friend operator < (Cow a,Cow b){return a.start < b.start;}
}cow[50005];
struct Stall{int No,end;bool friend operator < (Stall a,Stall b){return a.end > b.end;}Stall (int e,int n):end(e),No(n){}
};
int main(){int n,total,cnt,pos[50005];priority_queue <Stall> pq;scanf("%d",&n);for(int i = 0;i < n;i++){scanf("%d%d",&cow[i].start,&cow[i].end);cow[i].No = i;//记录奶牛的编号,因为只要一进行排序,奶牛本身的编号就会乱掉}sort(cow,cow + n);total = 0;for(int i = 0;i < n;i++){if(pq.empty()){pos[cow[i].No] = ++total;pq.push(Stall(cow[i].end,total));//记录栅栏的编号,同时记录栅栏的使用结束时间}else{Stall st = pq.top();if(st.end < cow[i].start){pq.pop();pos[cow[i].No] = st.No;pq.push(Stall(cow[i].end,st.No));}else{pos[cow[i].No] = ++total;pq.push(Stall(cow[i].end,total));}}}printf("%d\n",total);for(int i = 0;i < n;i++)printf("%d\n",pos[i]);return 0;
}