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
4#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
struct Cow{
int a,b;
int no;
bool operator <(const Cow &c) const{
return a<c.a;//根据牛挤奶的开始时间从小到大排序
}
}cows[50001];
int pos[50001];
struct stoll{
int ed;
int no;
bool operator <(const stoll &s) const{
return ed>s.ed;//在优先队列中，此表示最小的在队列前面
}
};
priority_queue <stoll> q;
int main(){
int n;
int num=0;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d%d",&cows[i].a,&cows[i].b);
cows[i].no=i;
}
sort(cows,cows+n);
int cnt=0;
for(int i=0;i<n;i++){
if(q.empty()){//若队列为空，则直接分配栅栏
cnt++;
stoll s;
s.ed=cows[i].b;
s.no=cnt;
q.push(s);
pos[cows[i].no]=cnt;
}
else{
stoll st=q.top();
if(st.ed<cows[i].a){//若此刻的奶牛挤奶的开始时间大于队列中结束时间最短的（说明直接可以进入该栅栏，无需分配其他）
q.pop();//删除上头牛的记录
stoll s;
s.ed=cows[i].b;
s.no=st.no;
q.push(s);//把此刻占用栅栏的牛的信息放入队列
pos[cows[i].no]=s.no;
}
else{
cnt++;
stoll s;
s.ed=cows[i].b;
s.no=cnt;
q.push(s);
pos[cows[i].no]=cnt;
}
}
}
printf("%d\n",cnt);
for(int i=0;i<n;i++){
printf("%d\n",pos[i]);
}
}