题目链接：
思路：
首先根据挤奶时间的先后顺序排序。。。然后将第一头牛加入优先队列。。然后就是加入优先队列的牛应该根据越早结束挤奶那么优先级更高，如果时间结束点相等，那么开始时间早的优先级高。。。
然后从前向后枚举。如果碰到有牛的挤奶时间的开始值大于优先队列的首部的结束值，那么说明这两头牛可以一起公用一个挤奶房。。然后从优先队列中删除这头牛。。那么这个问题就得到解决了。。。
题目：
|
Language: Default 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 Help FJ by determining:
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 |
||||||||||
代码为：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=50000+10;
int order[maxn];
struct Node
{
int st,en,pos;
friend bool operator<(Node a,Node b)
{
if(a.en==b.en)
return a.st<b.st;
return a.en>b.en;
}
}node[maxn];
bool cmp(Node a,Node b)
{
if(a.st==b.st)
return a.en<b.en;
else
return a.st<b.st;
}
priority_queue<Node>Q;
int main()
{
int n,ans;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
scanf("%d%d",&node[i].st,&node[i].en);
node[i].pos=i;
}
sort(node+1,node+1+n,cmp);
ans=1;
Q.push(node[1]);
order[node[1].pos]=1;
for(int i=2;i<=n;i++)
{
if(!Q.empty()&&Q.top().en<node[i].st)
{
order[node[i].pos]=order[Q.top().pos];
Q.pop();
}
else
{
ans++;
order[node[i].pos]=ans;
}
Q.push(node[i]);
}
printf("%d\n",ans);
for(int i=1;i<=n;i++)
printf("%d\n",order[i]);
while(!Q.empty()) Q.pop();
}
return 0;
}
poj3190Stall Reservations(贪心+优先队列),布布扣,bubuko.com