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poj3190Stall Reservations(贪心+优先队列)

题目链接： 啊哈哈，点我点我
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
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 2849 Accepted: 1021 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:

Time123456789 10Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..Stall 3 .. .. c3>>>>>>>>> .. .. .. ..Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible. 【poj3190Stall Reservations(贪心+优先队列)】Source
USACO 2006 February Silver

代码为：
#include #include #include #include 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.stb.en; } }node[maxn]; bool cmp(Node a,Node b) { if(a.st==b.st) return a.enQ; 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



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