ＰＯＪ　３１９０　Stall Reservations　贪心＋优先队列

Stall Reservations

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10476		Accepted: 3702		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.

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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

1 10

2 4

3 6

5 8

4 7

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

算法分析：

题意：

n头母牛都有固定的挤奶时间段，问至少需要多少挤奶站才能让n头母牛挤奶成功。

分析：

因为每一头母牛都有固定的挤奶时间，第一，时间开始得早的母牛肯定先挤奶，所以我们按照开始时间对其排序。第二，时间结束的早的母牛可以让出挤奶台，所以我们将正在挤奶的母牛压入一个优先队列（结束时间早在队首）。这样循环放入，得到最优解。

思想：贪心+优先队列

代码实现：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
using namespace std;  
const double eps = 1e-8;  
typedef long long LL;  
typedef unsigned long long ULL;  
const int INT_INF = 0x3f3f3f3f;  
const int INT_M_INF = 0x7f7f7f7f;  
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;  
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};  
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};  
const int MOD = 1e9 + 7;  
const double pi = acos(-1.0);  
const int MAXN = 1e5 + 10;  
const int MAXT = 10000 + 10;  
const int M=50010;
struct C
{
	int p,l,r;
	bool operator<(const C &c)const
	{
	  if(r==c.r)
	   return l>c.l;
	  else 
	  return r>c.r;
	}
};
bool cmp(const C &a,const C &b)
{
	if(a.l!=b.l)
	return a.l<b.l;
	else 
	return a.r<b.r;
	
}
int main()
{
	int n;
	while (scanf("%d",&n)!=EOF)
	{
		C c[M];
		priority_queue<C> q;     //结束时间早在队首
		int u[M];                 //表示i个机器用第几个摊位
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&c[i].l,&c[i].r);
			 c[i].p=i+1;
		}
		sort(c,c+n,cmp);   //开始时间从小到大排序
		q.push(c[0]);
		u[c[0].p]=1;
		int ans=1;
		for(int i=1;i<n;i++)
		{
			if(!q.empty()&&q.top().r<c[i].l)  //可以加入当前摊位
			{
				u[c[i].p]=u[q.top().p];
				q.pop();
			}
			else                              //需要另外添加摊位
			{
				ans++;
				u[c[i].p]=ans;
			}
			q.push(c[i]);
		}
		printf("%d\n",ans);
		for(int i=1;i<=n;i++)
			printf("%d\n",u[i]);
		
	}
    return 0;
}

ＰＯＪ ３１９０ Stall Reservations 贪心＋优先队列

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ＰＯＪ　３１９０　Stall Reservations　贪心＋优先队列