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POJ 3190 Stall Reservations(贪心+优先队列优化）

2024-07-14 23:54:32   225人阅读

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.

Source

USACO 2006 February Silver

额，贪心题=。=，STL里的优先队列

贪心策略：对奶牛挤奶开始时间由大到小排序，这样保证不会考虑有奶牛的开始挤奶时间在前面；

对入队的元素按挤奶结束时间时间由小到大排序，这样保证第一次就可以判定用不用添加槽=。=

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<queue>
using namespace std;
const int maxn=1000000+10;
struct node{
    int st,ed;
    int pos;//pos记录是第几个奶牛
    bool operator<(const node &a)const
    {
        if(ed==a.ed)
           return st>a.st;
        return ed>a.ed;
    }
}a[maxn];
int used[maxn];//used记录奶牛挤奶的位置=。=
int cmp(node l1,node l2)
{
    if(l1.st==l2.st)
        return l1.ed<l2.ed;
    return l1.st<l2.st;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        priority_queue<node>q;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&a[i].st,&a[i].ed);
            a[i].pos=i;
        }
        sort(a+1,a+n+1,cmp);
        q.push(a[1]);
        int ans=1;
        used[a[1].pos]=1;
        for(int i=2;i<=n;i++)
        {
            if(!q.empty()&&q.top().ed<a[i].st)//判定是否符合条件
            {
                used[a[i].pos]=used[q.top().pos];
                q.pop();
            }
            else//不符合条件槽的数量加一，同时这个奶牛应该在新加的槽
            {
                ans++;
                used[a[i].pos]=ans;
            }
            q.push(a[i]);
        }
        printf("%d\n",ans);
        for(int i=1;i<=n;i++)
            printf("%d\n",used[i]);
    }
    return 0;
}



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