poj 3190 Stall Reservations 贪心

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.

题意：

一些奶牛要在指定的时间内挤牛奶，而一个机器只能同时对一个奶牛工作。给你每头奶牛的指定时间的区间，问你最小需要多少机器，并且输出每头奶牛所用的机器序号。

按照每头奶牛挤牛奶的开始时间排序后，维护一个优先队列，按照奶牛的结束时间排序从小到大排序，每次比较队首元素中奶牛的结束时间和当前奶牛的开始时间，若大，则需要一台新机器，否则用队首的那台机器

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
struct p
{
    int x;
    int y;
    int z;
    friend bool operator<(p n1,p n2)//维护一个优先队列
    {
        return n1.y>n2.y;
    }
} a[50010];
bool cmp(p x,p y)
{
    if(x.x==y.x)
        return x.y<y.y;
    return x.x<y.x;//按照开始时间从小到大排序
}
int main()
{
    int t;
    int s[50010];
    while(~scanf("%d",&t))
    {
        memset(s,0,sizeof(s));
        for(int i=0; i<t; i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            a[i].z=i;//记录位置
        }
        sort(a,a+t,cmp);
        priority_queue<p>Q;
        Q.push(a[0]);
        s[a[0].z]=1;//记录使用的机器序号
        int sum=2;
        for(int i=1; i<t; i++)
        {
            if(Q.top().y>=a[i].x)
                s[a[i].z]=sum++;
            else
            {
                s[a[i].z]=s[Q.top().z];
                Q.pop();
            }
            Q.push(a[i]);
        }
        printf("%d\n",sum-1);
        for(int i=0; i<t; i++)
            printf("%d\n",s[i]);
    }
    return 0;
}

poj 3190 Stall Reservations 贪心

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