Stall Reservations POJ - 3190 （贪心+优先队列）

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11002		Accepted: 3886		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：人去挤奶牛，一个人挤一头，输出最少几人及牛奶并且哪只牛哪个人

题解：贪心加优先队列，将开始时间在前面的排在前面，开始时间一样的就把结束时间后的排在后面

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=;
int n,use[maxn];

struct node
{
    int st;
    int en;
    int pos;
    bool operator < (const node &a)const
    {
        if(en==a.en)
            return st>a.st;
        return en>a.en;
    }
}a[maxn];
priority_queue<node>q;
bool cmp(node a,node b)
{
    if(a.st==b.st)
        return a.en<b.en;     //排序，按照开始时间排序，开始时间相同的结束时间迟的排后面
    return a.st<b.st;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=;i<n;i++)
        {
            scanf("%d %d",&a[i].st,&a[i].en);
            a[i].pos=i;
        }
        sort(a,a+n,cmp);
        q.push(a[]);
        int ans=;
        use[a[].pos]=;
        for(int i=;i<n;i++)
        {
            if(!q.empty() && q.top().en < a[i].st)
            {
                use[a[i].pos]=use[q.top().pos];
                q.pop();
            }
            else
            {
                ans++;
                use[a[i].pos]=ans;
            }
            q.push(a[i]);
        }
        cout<<ans<<endl;
        for(int i=;i<n;i++)
            cout<<use[i]<<endl;
        while(!q.empty())
            q.pop();
    }
    return ;
}