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poj3190Stall Reservations(贪心+优先队列)

2024-07-15 09:11:08   222人阅读

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:
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: 

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

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;
}



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