PKUoj 3190 Stall Reservations（贪心算法）

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10799		Accepted: 3818		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

源代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
struct cow{
    int a;
    int b;
    int id;
    bool operator <(const cow &c) const{
        return a<c.a;
    }
}cows[50000+5];
int pos[50000+5];//表示编号为i的奶牛去的畜栏编号
struct Stall{
    int e;//设置畜栏结束使用时间，也就是奶牛挤完奶的时间
    int id;//设置畜栏的编号
    bool operator<(const Stall &s)const{
        return e>s.e;
        //结束时间越早的优先级越高
    }
    Stall(int e,int n):e(e),id(n){}
};
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].id=i;//奶牛的编号就是变量i
    }
    sort(cows,cows+n);
    int total=0;//设置栅栏计数
    priority_queue<Stall>pq;//设置畜栏队列pq

    for(int i=0;i<n;++i){
        if(pq.empty()){//畜栏为空的情况，对应i等于的情况
            ++total;
            pq.push(Stall(cows[i].b,total));//将奶牛结束时间，畜栏数目压入对列
            pos[cows[i].id]=total;
        }
        else{//对应i不为0的情况
            Stall st =pq.top();//取顶操作
        if(st.e<cows[i].a){//端点也不能重合
            pq.pop();//弹出操作
            pos[cows[i].id]=st.id;//记录畜栏id
            pq.push(Stall(cows[i].b,st.id));//不断更新畜栏结束时间还有队顶id
            }
        else{//找不到空的畜栏
            ++total;
        pq.push(Stall(cows[i].b,total));
        pos[cows[i].id]=total;
        }
        }
    }
    printf("%d\n",total);
    for(int i=0;i<n;i++)
        printf("%d\n",pos[i]);
    return 0;
}

思路：明显要先处理产奶时间靠前并且结束时间靠前的奶牛，这样才能使后面的奶牛接着使用同一畜栏。

这里就要使用到优先队列priority_queue。

操作步骤如下：

①所有奶牛按照开始时间从小到大排序。

②为第一头奶牛分配一个畜栏

③依次处理后面每头奶牛i，考虑已经分配到畜栏中，结束时间最早的畜栏x

使得结束时间最早的畜栏始终位于队列头部。

PKUoj 3190 Stall Reservations（贪心算法）

Stall Reservations

猜你喜欢