POJ-3190 Stall Reservations

题目传送门
Stall Reservations
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7741 Accepted: 2742 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

USACO 2006 February Silver

题意：
农夫有N头奶牛，每头奶牛都有自己的产奶时间段，时间start到时间end，每头牛产奶时必须单独为它安排一间畜栏，不能多头奶牛（大于等于2）同时共用一间，要求根据每头牛的产奶时间段，安排一个合理的方案使得使用畜栏数最少。

思路：
这个题，一看就知道是贪心，可是怎么贪心呢？我们也知道，一般这样的区间问题，基本上是排序问题，这个题目也是如此，说说我的思路吧，我首先将整个时间区间，按照时间start从小到大排序，将时间start最小的奶牛加入正在产奶的priority_queue中，priority_queue中定义时间end更小的优先级更高，现在已经有奶牛在产奶了，所以我们在排好序的时间区间中选择，和正在产奶的优先级最高的时间区间匹配的时间区间，也就是不相交；匹配，说明可以在同一个畜栏产奶，不匹配，就重新安排一个畜栏产奶。

我简单的理解为:正在产奶的结束时间越早越好，等待产奶的开始时间越早越好，这样进行匹配贪心得到最优解。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int maxn=5*1e4+100;int cnt[maxn];struct Node{    int start,end,id;    bool operator <(const Node &a)const    {        return a.end<end;    }}node[maxn];bool cmp(Node a,Node b){    return a.start<b.start;}int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            scanf("%d%d",&node[i].start,&node[i].end);            node[i].id=i;        }        sort(node,node+n,cmp);        priority_queue<Node>que;        que.push(node[0]);         cnt[node[0].id]=1;        int ans=1;        for(int i=1;i<n;i++)        {            if(!que.empty()&&que.top().end<node[i].start)            {                cnt[node[i].id]=cnt[que.top().id];                que.pop();            }            else            {                cnt[node[i].id]=++ans;            }            que.push(node[i]);        }        printf("%d\n",ans);        for(int i=0;i<n;i++)        {            printf("%d\n",cnt[i]);        }    }    return 0;}