poj 3190 Stall Reservations（贪心）

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.

题意：奶牛要在指定的时间挤奶，一台机器只能给一头奶牛挤奶，问最少要几台机器。

思路：用STL里的优先队列，判断空闲机器是否满足下一头奶牛的挤奶时间，满足就让挤完奶的奶牛出队把机器编号给下一头奶牛，下一头奶牛入队，否则机器+1，下一头奶牛入队.

注意：输出的顺序和输入的顺序相同，在输入的时候把下标一起存入结构体，另外开一个数组记录机器的编号，在排序后下标就不会乱了。

ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
struct node
{
    int a;
    int b;
    int c;
    bool operator<(const node &c)const
    {
        if(b==c.b)
            return a>c.a;
        return b>c.b;
    }
} d[60000];
priority_queue<node> q;
int s[60000];
bool cmp(struct node a,struct node b)
{
    if(a.a==b.a)
        return a.b<b.b;
    return a.a<b.a;
}
int main()
{
    int a,b,i,j,n=1;
    scanf("%d",&a);
    for(i=1; i<=a; i++)
    {
        scanf("%d%d",&d[i].a,&d[i].b);
        d[i].c=i;
    }
    sort(d+1,d+a+1,cmp);
    q.push(d[1]);
    s[d[1].c]=1;
    for(i=2; i<=a; i++)
    {
        int x=q.top().b;
        if(x<d[i].a)
        {
            s[d[i].c]=s[q.top().c];
            q.pop();
        }
        else
        {
            n++;
            s[d[i].c]=n;
        }
        q.push(d[i]);
    }
    printf("%d\n",n);
    for(i=1; i<=a; i++)
        printf("%d\n",s[i]);
}

 

 有什么问题可以评论提问。