Poj 3190 Stall Reservations【贪心+优先队列】

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4746		Accepted: 1694		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.

题意：

某个人有n头奶牛，给出每只奶牛产奶的开始时间和结束时间，问这个人最少需要多少个房子才能使得每只奶牛（好像都是母牛...）产奶的时候都可以单独使用一个房子

输出最少需要的房子的数量，并且按给出的奶牛顺序输出每只奶牛使用的房子的编号（错了两次才发现这个条件！！）

题解：

区间选择问题，贪心的思想

步骤如下：

1，奶牛是需要排序的，但是输出的顺序必须是按给出的顺序，因此需要给每只奶牛编号，用一个数组（num）映射来保存他的位置信息

2，为了使得尽量使得区间不相交，对奶牛排序，一级排序按开始时间递增（最先开始的优先选择），二级排序按结束时间递增

3，使用优先队列，队列中的元素代表当前在产奶的奶牛，队列中的元素结束时间的较小者优先出队，保证尽量空出来多的房子

4，当新加入一只奶牛时，如果队首奶牛还没有产奶结束，只能新使用一个房子，否则就使用当前队首的奶牛使用的房子，并且把队首的奶牛出队（产奶结束）

5，遍历所有奶牛，并记录奶牛的编号和房子的编号的对应关系，最终输出房子的编号即为所求

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
struct cow
{
int bg,ed,num;
bool friend operator <(cow a,cow b)
{
return a.ed>b.ed;
}
}x[100005];

bool cmp(cow a,cow b)
{
if(a.bg==b.bg)
{
a.ed<b.ed;//最早结束的
}
return a.bg<b.bg;
}
int main()
{
int n,num[50005];
//freopen("shuju.txt","r",stdin);
while(~scanf("%d",&n))
{
priority_queue<cow> q;
for(int i=0;i<n;++i)
{
scanf("%d%d",&x[i].bg,&x[i].ed);
x[i].num=i;//牛编上号
}
sort(x,x+n,cmp);//排序
int cnt=0;
num[x[0].num]=cnt++;//第一头牛的编号
q.push(x[0]);
for(int i=1;i<n;++i)
{
cow st=q.top();
if(x[i].bg>st.ed)//有头牛的时间结束
{
q.pop();
num[x[i].num]=num[st.num];//下头牛 使用这个编号
}
else//当前没有结束
{
num[x[i].num]=cnt++;
}
q.push(x[i]);
}
printf("%d\n",cnt);
for(int i=0;i<n;++i)
{
printf("%d\n",num[i]+1);
}
}
return 0;
}