Stall Reservations

题目链接http://poj.org/problem?id=3190

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:10223		Accepted:3608		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.

需要用到贪心思想的模拟题。难点是维护每一个畜栏的结束时间。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#define DEBUG(x) cout<<#x<<" = "<<x<<endl
using namespace std;
const int MAXN=5e4+100;
struct cow{
    int s,e;
    int idx;
    cow(){}
    cow(int ss,int ee,int i):s(ss),e(ee),idx(i){}
    bool operator<(const cow &c)const
    {
        return s<c.s;
    }
};
struct stall{
    int et;
    int no;
    stall(int e,int n):et(e),no(n){}
    stall(){}
    bool operator<(const stall &s)const
    {
        return et>s.et;
    }
};
int N;
int pos[MAXN];
cow cowarray[MAXN];
int main()
{
//    freopen("in.txt","r",stdin);
    cin>>N;
    for(int i=0;i<N ;i++ ){
        int s,e;
        scanf("%d%d",&s,&e);
        cowarray[i].s=s,
        cowarray[i].e=e,
        cowarray[i].idx=i;
    }
    sort(cowarray,cowarray+N);
    priority_queue<stall>pq;
    int cnt=1;
    cow &c=cowarray[0];
    pq.push(stall(c.e,cnt));
    ///这是一个经常出现的bug，且难以发现，排序之后的下标和之前的下标不同
//    pos[0]=cnt;
    pos[c.idx]=cnt;
    for(int i=1;i<N ;i++ ){
        stall st=pq.top();
        if(st.et<cowarray[i].s){
            pq.pop();
            st.et=cowarray[i].e;
            int idx=cowarray[i].idx;
            pos[idx]=st.no;
            pq.push(st);
        }
        else{
            cnt++;
            cow &t=cowarray[i];
            pq.push(stall(t.e,cnt));
            pos[t.idx]=cnt;
        }
    }
    printf("%d
",cnt);
    for(int i=0;i<N ;i++ ){
        printf("%d
",pos[i]);
//        cout<<pos[i]<<endl;
    }
}