POJ 3190 Stall Reservations （优先队列+结构体）

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

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: 

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.

 

题意：题目意思就是每个谷仓同时只能有一头牛在里面挤奶，每头牛有个开始和结束时间。

问你最多要多少谷仓让所有牛挤完奶，有个细节就是如果这头牛开始时间等于进谷仓的上头牛的结束时间，

这样算交集，不能连起来。

思路：一眼贪心题，用优先队列维护每个谷仓的结束时间，结束时间早的在队首，先出队。如果有未进谷仓牛的

开始时间大于这个结束时间，那么把这个结束时间出队，将这头牛的结束时间入队。

考虑到题目要输出每头牛按序号的进了哪个谷仓，那我们就要记录一下谷仓，最后按序号排序后输出。

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<list>
#include<stack>
//#include<unordered_map>
using namespace std;
#define ll long long 
#define dd cout<<endl
const int mod=1e9+7;
const int nf=1e9+7;

const int maxn=5e4+10;

typedef struct
{
    int a;
    int b;
    int number;
} St;
//优先队列+结构体，定义优先级 
bool operator<(const St &x,const St &y)
{
    return x.b > y.b;
 } 
 
bool cmp(const St &x,const St &y)
{
    if(x.a != y.a)
        return x.a < y.a;
    else
        return x.b < y.b;
}
 
St sum[maxn];

int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    
    int n;
    cin>>n;
    
    int a,b;
    
    priority_queue<St>q;    
    
    int ans=0;
    vector<St>v;

    for(int i=0;i<n;i++)
    {
        cin>>sum[i].a>>sum[i].b;
        sum[i].number=i;
    }
    
    sort(sum,sum+n,cmp);//这里先根据开始顺序排下序 
    
    for(int i=0;i<n;i++)
    {
        a=sum[i].a;//当前开始时间 
        b=sum[i].b;//当前结束时间 
        
        if(q.empty())//空的直接进篱笆 
        {
            ans++;
            q.push({ans,b});
            v.push_back({sum[i].number,ans});//记录 
        }
        else
        {
            St now=q.top();
            
            if(now.b < a)//当前开始时间大于最早的结束时间 
            {
                q.pop();
                
                v.push_back({sum[i].number,now.a});//记录 
                q.push({now.a,b});
            }
            else//在进一个篱笆 
            {
                ans++;
                q.push({ans,b});
                v.push_back({sum[i].number,ans});//记录 
            }
        }    
    }
    
    cout<<ans<<endl;//篱笆数 
    
    sort(v.begin(),v.end(),cmp);//按奶牛序号排序 
    
    for(int i=0;i<v.size();i++)//输出答案 
        cout<<v[i].b<<endl;
    
    return 0;
}