POJ_1456 Supermarket

Supermarket

链接

POJ_1456 Supermarket

Description

A supermarket has a set Prod of products on sale. It earns a profit $p_x$ for each product $x\in Prod$ sold by a deadline $d_x$ that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products $Sell\le Prod$ such that the selling of each product $x\in Sell$, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is $Profit(Sell)={\sum }_{x\in Sell}px$. An optimal selling schedule is a schedule with a maximum profit.

For example, consider the products P r o d = { a , b , c , d } Prod=\left\{a,b,c,d\right\} Prod={a,b,c,d} with $\left(p_a,d_a\right)=\left(50,2\right)$,$\left(p_b,d_b\right)=\left(10,1\right)$, $\left(p_c,d_c\right)=\left(20,2\right)$, and $\left(p_d,d_d\right)=\left(30,1\right)$. The possible selling schedules are listed in table $1$. For instance, the schedule S e l l = { d , a } Sell=\left\{d,a\right\} Sell={d,a} shows that the selling of product $d$ starts at time $0$ and ends at time $1$, while the selling of product $a$ starts at time $1$ and ends at time $2$. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is $80$.

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.

题目大意：超市里有$N$件商品，每件商品都有利润$p_i$和过期时间$d_i$，每天只能卖一件商品，过期商品不能再卖。

求合理安排每天卖的商品的情况下，可以得到的最大收益是多少。

Input

A set of products starts with an integer $0\le n\le 10000$, which is the number of products in the set, and continues with n pairs pi di of integers, $1\le p_i\le 10000$ and $1\le d_i\le 10000$, that designate the profit and the selling deadline of the $i$-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

Output

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

Sample Input

4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8 2
   5 20  50 10

Sample Output

80
185

Hint

The sample input contains two product sets. The first set encodes the products from table $1$. The second set is for $7$ products. The profit of an optimal schedule for these products is $185$.

思路

先建立一个小根堆。

把所有的产品按照过期时间排序，然后遍历每一件产品。

若某产品的过期时间等于堆中的产品数，即这件产品如果再不卖就会过期，那么将它与堆顶价值最小的产品比较。如果该产品的价值比堆顶产品的价值还高，说明此方案更优。这时就要删除堆顶，将该产品的价值加入堆。

若某产品的过期时间大于堆中的产品数，即这件产品在堆中所有产品卖完后依然不会过期。这时就要直接将该产品加入堆。

最后堆中所有产品的价值之和即位所求答案。

这是第一次做堆的题目，所以堆是手写的。

代码

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
const ll N=10000;
struct str{
	ll p,d;
}prod[N+1];
ll n,h_cnt,ans,heap[N+1];
bool cmp(str a,str b)
{
	return a.d<b.d;
}
void up(ll p)//将堆中的某一个元素上移直至满足小根堆的性质
{
	while(p>1)
	{
		if(heap[p]<heap[p/2])//子节点<父节点
		{
			swap(heap[p],heap[p/2]);
			p=p>>1;
		}
		else
			break;
	}
	return;
}
void down(ll p)//将堆中的某一个元素下沉直至满足小根堆的性质
{
	ll s=p*2;//s是p的儿子 
	while(s<=h_cnt)
	{
		if(s<h_cnt&&heap[s]>heap[s+1])//取左右儿子中较小的那个 
			s++;
		if(heap[s]<heap[p])
		{
			swap(heap[s],heap[p]);
			p=s,s=p<<1;
		}
		else
			break;
	}
	return;
}
void h_insert(ll val)//向堆中插入一个带有权值的节点 
{
	h_cnt++;
	heap[h_cnt]=val;
	up(h_cnt);
	return;
}
void h_extract()//删除堆顶节点 
{
	heap[1]=heap[h_cnt];
	heap[h_cnt]=0;
	h_cnt--;
	down(1);
	return;
}
int main()
{
	//freopen("test.in","r",stdin);
	//freopen("test.out","w",stdout);
	while(cin>>n)//这里只能写cin，否则会无法判断是否有输入
	{
		h_cnt=0;
		memset(heap,0,sizeof(heap));
		for(ll i=1;i<=n;i++)
			scanf("%lld%lld",&prod[i].p,&prod[i].d);
		sort(prod+1,prod+n+1,cmp);
		for(ll i=1;i<=n;i++)
		{
			if(prod[i].d==h_cnt&&heap[1]<prod[i].p)//将要过期且比堆顶更优 
			{
				h_extract();//移除堆顶
				h_insert(prod[i].p);//将当前产品的价值插入堆 
			}
			if(prod[i].d>h_cnt)//不可能过期
				h_insert(prod[i].p); 
		}
		ans=0;
		while(h_cnt)
		{
			ans=ans+heap[1];
			h_extract();
		}
		printf("%lld\n",ans);
	}
	return 0;
}