Stall Reservations（POJ 3190）

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10784		Accepted: 3813		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

解题思路：

奶牛到挤奶的时间就必须要挤，只需逐个观察该奶牛需要挤奶时，有没有空牛棚即可。

1）把所有奶牛根据开始时间从小到大进行排序。

2）为第一头奶牛分配一个牛棚。

3）依次对后面的奶牛进行操作，处理第 i 头时，考虑这时有没有空牛棚（上一只奶牛挤奶结束）。

令结束时间最早的牛棚结束时间为E(x)（奶牛挤奶结束时间），S（i）为第i头奶牛开始的时间

如果E（x） < S（i）则不用新增牛棚，i 可以进入 x，并修改E（i）。

如果E（x） >= S（i）则需要新增牛棚，并使该牛棚的结束事假为E（i）。

在为牛分配牛棚时存在三种情况

1）还没有建造过牛棚，此时就要新增牛棚，并把牛棚结束时间改为第一头牛的E（i）。

2）建过牛棚，该牛可以进入最早结束的牛棚进行操作。

3）建过牛棚，但是最早结束的牛棚的结束时间 >= 该牛的开始时间,则要新增牛棚。

注意：优先队列默认为从大到小排列，如果要从小到大，排列规则的符号与一般的相反（> 改成 <）

代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
struct cow {
    int a,b;
    int nu;   //牛的编号
    bool operator < (const cow & c) const{
        return a < c.a;
    }
}cows[50000+10];
int pos [50000+10]; //pos[i]表示编号为i的牛去的牛棚
struct stall{
    int end;
    int nu;
    bool operator <(const stall & c) const{
        return end > c.end;
    }
    stall(int x,int y):end(x),nu(y){ }
};
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 0;i < n;i++){
        scanf("%d %d",&cows[i].a,&cows[i].b);
        cows[i].nu = i;
    }
    sort(cows,cows+n);
    int total = 0;
    priority_queue<stall> pq;
    for(int i = 0;i < n;i++){
        if(pq.empty()){
            total++;
            pq.push (stall(cows[i].b,total));
            pos[cows[i].nu] = total;
        }
        else{
            stall st = pq.top();
            if(st.end < cows[i].a){
                pq.pop();
                pos[cows[i].nu] = st.nu;
                pq.push (stall(cows[i].b,st.nu));
            }
            else{
                total++;
                pos[cows[i].nu] = total;
                pq.push (stall(cows[i].b,total));
            }
        }
    }
    printf("%d\n",total);
    for(int i = 0;i < n;i++){
        printf("%d\n",pos[i]);
    }
    return 0;
}

Stall Reservations（POJ 3190）

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