Stall Reservations
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 11002 | Accepted: 3886 | Special Judge | ||
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.
Source
题意：人去挤奶牛，一个人挤一头，输出最少几人及牛奶并且哪只牛哪个人
题解：贪心加优先队列，将开始时间在前面的排在前面，开始时间一样的就把结束时间后的排在后面
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=;
int n,use[maxn]; struct node
{
int st;
int en;
int pos;
bool operator < (const node &a)const
{
if(en==a.en)
return st>a.st;
return en>a.en;
}
}a[maxn];
priority_queue<node>q;
bool cmp(node a,node b)
{
if(a.st==b.st)
return a.en<b.en; //排序，按照开始时间排序，开始时间相同的结束时间迟的排后面
return a.st<b.st;
} int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=;i<n;i++)
{
scanf("%d %d",&a[i].st,&a[i].en);
a[i].pos=i;
}
sort(a,a+n,cmp);
q.push(a[]);
int ans=;
use[a[].pos]=;
for(int i=;i<n;i++)
{
if(!q.empty() && q.top().en < a[i].st)
{
use[a[i].pos]=use[q.top().pos];
q.pop();
}
else
{
ans++;
use[a[i].pos]=ans;
}
q.push(a[i]);
}
cout<<ans<<endl;
for(int i=;i<n;i++)
cout<<use[i]<<endl;
while(!q.empty())
q.pop();
}
return ;
}