POJ 3190 Stall Reservations（贪心）

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最新推荐文章于 2022-03-25 20:28:13 发布

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本文介绍了一种算法，用于解决多个奶牛在特定时间内挤奶的问题，确保每头奶牛都能在不冲突的时间段内使用挤奶位。通过有效的资源分配策略，实现了最小挤奶位数量的需求。

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Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3590		Accepted: 1284		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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define epst 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 50005

struct stud{
  int x,y;
  int pos;
  bool operator <(const stud d) const
  {
		return y>d.y;
  }
}f[N];

int ans[N];
int k;
priority_queue<stud>q;
int n;

int cmp(stud a,stud b)
{
	if(a.x==b.x) return a.y<b.y;
	return a.x<b.x;
}

void solve()
{
	int i,j;
	stud cur;
	while(!q.empty())q.pop();
	for(i=1;i<=n;i++)
	{
		if(q.empty())
		{
			ans[f[i].pos]=++k;
			q.push(f[i]);
			continue;
		}
        cur=q.top();
        if(cur.y<f[i].x)
		{
			ans[f[i].pos]=ans[cur.pos];
			q.pop();
		}
        else  ans[f[i].pos]=++k;
        q.push(f[i]);
	}
}

int main()
{
	int i,j;
	while(~sf(n))
	{
		for(i=1;i<=n;i++)
			{
				sff(f[i].x,f[i].y);
			    f[i].pos=i;
			}
		sort(f+1,f+n+1,cmp);
		k=0;
		solve();
		pf("%d\n",k);
		for(i=1;i<=n;i++)
			pf("%d\n",ans[i]);
	}
    return 0;
}