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

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

Sample Output

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。

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<cmath>
  5 #include<algorithm>
  6 #include<map>
  7 #include<set>
  8 #include<vector>
  9 #include<queue>
 10 #include<list>
 11 #include<stack>
 12 //#include<unordered_map>
 13 using namespace std;
 14 #define ll long long 
 15 #define dd cout<<endl
 16 const int mod=1e9+7;
 17 const int nf=1e9+7;
 18 
 19 const int maxn=5e4+10;
 20 
 21 typedef struct
 22 {
 23     int a;
 24     int b;
 25     int number;
 26 } St;
 27 //优先队列+结构体，定义优先级 
 28 bool operator<(const St &x,const St &y)
 29 {
 30     return x.b > y.b;
 31  } 
 32  
 33 bool cmp(const St &x,const St &y)
 34 {
 35     if(x.a != y.a)
 36         return x.a < y.a;
 37     else
 38         return x.b < y.b;
 39 }
 40  
 41 St sum[maxn];
 42 
 43 int main()
 44 {
 45     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 46     
 47     int n;
 48     cin>>n;
 49     
 50     int a,b;
 51     
 52     priority_queue<St>q;    
 53     
 54     int ans=0;
 55     vector<St>v;
 56 
 57     for(int i=0;i<n;i++)
 58     {
 59         cin>>sum[i].a>>sum[i].b;
 60         sum[i].number=i;
 61     }
 62     
 63     sort(sum,sum+n,cmp);//这里先根据开始顺序排下序 
 64     
 65     for(int i=0;i<n;i++)
 66     {
 67         a=sum[i].a;//当前开始时间 
 68         b=sum[i].b;//当前结束时间 
 69         
 70         if(q.empty())//空的直接进篱笆 
 71         {
 72             ans++;
 73             q.push({ans,b});
 74             v.push_back({sum[i].number,ans});//记录 
 75         }
 76         else
 77         {
 78             St now=q.top();
 79             
 80             if(now.b < a)//当前开始时间大于最早的结束时间 
 81             {
 82                 q.pop();
 83                 
 84                 v.push_back({sum[i].number,now.a});//记录 
 85                 q.push({now.a,b});
 86             }
 87             else//在进一个篱笆 
 88             {
 89                 ans++;
 90                 q.push({ans,b});
 91                 v.push_back({sum[i].number,ans});//记录 
 92             }
 93         }    
 94     }
 95     
 96     cout<<ans<<endl;//篱笆数 
 97     
 98     sort(v.begin(),v.end(),cmp);//按奶牛序号排序 
 99     
100     for(int i=0;i<v.size();i++)//输出答案 
101         cout<<v[i].b<<endl;
102     
103     return 0;
104 }

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

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