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

Sample Output

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=60000;
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=0;i<n;i++)
        {
            scanf("%d %d",&a[i].st,&a[i].en);
            a[i].pos=i;
        }
        sort(a,a+n,cmp);
        q.push(a[0]);
        int ans=1;
        use[a[0].pos]=1;
        for(int i=1;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=0;i<n;i++)
            cout<<use[i]<<endl;
        while(!q.empty())
            q.pop();
    }
    return 0;
}

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

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