Stall Reservations POJ - 3190（贪心）

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:N头牛，每头牛都有霸占食槽的习惯，从【cow【i】.l,cow【i】.r】，被霸占的食槽，无法给其他牛使用，就必须多几个平行食槽

问最少几个平行食槽可供牛使用，且输出每个牛所在食槽处。

 

思路：一个食槽，能否放入下一个牛，取决于前一个牛的cow【i-1】.r 是否小于 cow【i】.l 

我们可以想到，如果cow【i-1】.r >= cow【i】.l ,  就需要另外开一行食槽。

cow【i-1】.r < cow【i】.l 那我们就可以把当前的cow【i】放到这一行食槽中。 

但是这样不一定是最优的，你虽让能放进去，但是前面牛的越早结束越好。

 这样的话就可以用一个最小堆维护每列牛最右边的食槽，

 

开始我想到这，以为有多少食槽就要建立多少个堆，让后再去找它符合哪个队（显然这样不行）,其实只用一个堆就可以了，

一个堆的话如果右边界最小的你都不满足，那么其他的你肯定不满足

我们还需要事前将牛按照左边界排序，因为对于同一个最小的右边界，我们肯定希望塞入左边界最小的牛

 

 

 #include<cstdio>
 #include<iostream>
 #include<queue>
 #include<algorithm>
 using namespace std;

 const int maxn = 5e4+;
 int n;
 struct Node
 {
     int id;
     int l,r;
 };

 int mp[maxn];
 Node cow[maxn];
 bool cmp1(Node a,Node b)
 {
     return a.l < b.l;
 }

 bool operator<(Node a,Node b)
 {
     return a.r > b.r;
 }

 priority_queue<Node>que;
 int main()
 {
     scanf("%d",&n);
     for(int i=; i<=n; i++)
     {
         scanf("%d%d",&cow[i].l,&cow[i].r);
         cow[i].id = i;
     }
     sort(cow+,cow++n,cmp1);
     int k = ;
     for(int i=; i<=n; i++)
     {
         if(que.empty())
         {
             mp[cow[i].id] = k;
         }
         else if(que.top().r < cow[i].l)
         {

             mp[cow[i].id] = mp[que.top().id];
             que.pop();
         }
         else
             mp[cow[i].id] = ++k;
         que.push(cow[i]);
     }
     printf("%d\n",k);
     for(int i=; i<=n; i++)
     {
         printf("%d\n",mp[i]);
     }
 }