poj3190 Stall Reservations (贪心+优先队列)

Cleaning Shifts

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 2

Problem Description

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.

 

Input

* Line 1: Two space-separated integers: N and T <br> <br>* Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.

 

Output

* Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.

 

Sample Input

3 10 1 7 3 6 6 10

 

Sample Output

2

 

 

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.

 

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。。。

#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
", ans);
        for (int i = 1; i <= n; i++)
            printf("%d
", order[i]);
        while (!Q.empty())  Q.pop();
    }
    return 0;
}