poj 3190 Stall Reservations

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

1 10

2 4

3 6

5 8

4 7

Sample Output

#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
struct Cow{   //定义一个奶牛的结构体，包含奶牛的挤奶开始时间a,结束时间b,和奶牛的序号No
    int a;
    int b;
    int No;
bool operator < (const Cow c) const{  //重载< 以挤奶时间从早到晚排序
    return a<c.a;
    }
 }cow[50010];
int pos[50010];  //编号为i的奶牛的畜栏号
struct Stall{     //畜栏结构体，畜栏的使用只看结束时间，所以包括使用结束时间和畜栏序号
    int end;
    int No;
bool operator < (const Stall s) const{   //重载<  优先队列中 时间结束早的在前
return end>s.end;
    }
    Stall(int e,int n):end(e),No(n){}
 };
int main(){
    int n;
    cin>>n;
for(int i=1;i<=n;i++){
    cin>>cow[i].a>>cow[i].b;
    cow[i].No=i;
    }
    sort(cow,cow+n);  //排序
    int total=0;           //total为畜栏数，开始为0
priority_queue<Stall> pq;   //奶牛队列
for(int i=1;i<=n;i++){
        if(pq.empty()){   //如果队列为空，即一个也没有时
        total++;            //即第一个奶牛占第一个畜栏
    pq.push(Stall(cow[i].b,total));  //插入奶牛结束挤奶时间和畜栏编号
    pos[cow[i].No]=total;            //畜栏编号
        }
        else{
        Stall st=pq.top();             //队列不为空时，st为首元素
        if(st.end<cow[i].a){         //奶牛结束时间小于另一只开始时间
            pq.pop();                    //删除此奶牛信息，下一步更新
            pos[cow[i].No]=st.No;  //此时用同一个畜栏
            pq.push(Stall(cow[i].b,st.No));  //更新奶牛信息
        }
        else{                       //奶牛结束时间大于另一只开始时间
            total++;             //创建新畜栏
            pq.push(Stall(cow[i].b,total));     //插入奶牛信息
            pos[cow[i].No]=total;               //畜栏编号
           }
       }
   }
 cout<<total<<endl;                            //总畜栏数
 for(int i=1;i<=n;i++)
    cout<<pos[i]<<endl;                      //奶牛所在畜栏编号
 return 0;
}