[POJ 1456] Supermarket【贪心 / 贪心+优先队列 / 贪心+并查集】

  • 2018-01-04
  • 0
  • 0

Problem:

Time Limit: 2000MS Memory Limit: 65536K

Description

A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx 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 ≤ Prod such that the selling of each product x∈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)=Σx∈Sellpx. An optimal selling schedule is a schedule with a maximum profit.
For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule 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.

Input

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 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.

Source

Southeastern Europe 2003

Solution:

在网上找并查集的题,结果给了一道贪心题。。

这道题的贪心思路很好理解,就是先将商品按价格从大到小排序,再枚举每个商品,尝试将其放在能放的最大时间处即可。详见 Code #1

Code #1 的最坏时间复杂度为 O(Tn2),并不能通过本题,然而 POJ 的数据实在太水,抱着必 TLE 的决心交上去竟然 AC 了。。

但不能就此罢休。浏览了一下 [discuss] 发现这题可以用优先队列优化。但是这次要按截止时间从大到小排序,然后枚举每个时间点,将这个时间点截止的商品扔进优先队列里,再取 1 次最大值(每个时间点只能卖 1 样商品)即可。详见 Code #2

Code #2 的时间复杂度为 O(Tnlogn),这样才是正解。

突然又看到这道贪心题真的有并查集优化的方法,而且十分巧妙。可以发现 Code #1 复杂度的瓶颈是寻找“能放的最大时间”,而这个可以用并查集的特点进行优化。画图易知,每次节点 i 放过之后,将 fa[i] 设置为 i - 1,这样下次 find() 的时候就一定会跳过 i,再通过路径压缩做到快速寻找能放的最大时间。详见 Code #3

Code #3 的时间复杂度为 O(Tnα(n)) ≈ O(Tn),效率较 Code #2 提高不少(虽然在 POJ 的水数据面前差异不太明显)

Code #1: O(Tn2) [252K, 141MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

int n, ans;
bool vis[10005];

struct Product{
	int val, deadl;
	
	inline bool operator < (const Product &prod2) const {return val > prod2.val;}
} prod[10005];

int main(){
	while(scanf("%d", &n) != EOF){
		for(register int i = 1; i <= n; i++) scanf("%d%d", &prod[i].val, &prod[i].deadl);
		sort(prod + 1, prod + n + 1);
		memset(vis, 0, sizeof(vis)), ans = 0;
		for(register int i = 1; i <= n; i++)
			for(register int j = prod[i].deadl; j > 0; j--)
				if(!vis[j]){
					vis[j] = 1, ans += prod[i].val;
					break;
				}
		printf("%d\n", ans);
	}
	return 0;
}

Code #2: O(Tnlogn) [404K, 79MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

int n, ans;

struct Product{
	int val, deadl;
	
	inline bool operator < (const Product &prod2) const {return deadl > prod2.deadl;}
} prod[10005];

priority_queue<int> q;

inline void clearPQ(priority_queue<int> &q) {priority_queue<int> tmp; swap(tmp, q);}

int main(){
	while(scanf("%d", &n) != EOF){
		int maxDate = 0;
		for(register int i = 1; i <= n; i++){
			scanf("%d%d", &prod[i].val, &prod[i].deadl);
			maxDate = max(maxDate, prod[i].deadl);
		}
		sort(prod + 1, prod + n + 1);
		clearPQ(q), ans = 0;
		int prodId = 1;
		for(register int curDate = maxDate; curDate > 0; curDate--){
			for(; prodId <= n && prod[prodId].deadl >= curDate; prodId++) q.push(prod[prodId].val);
			if(!q.empty()) ans += q.top(), q.pop();
		}
		printf("%d\n", ans);
	}
	return 0;
}

Code #3: O(Tnα(n)) [284K, 63MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

int n, ans, maxDate;

struct Product{
	int val, deadl;
	
	inline bool operator < (const Product &prod2) const {return val > prod2.val;}
} prod[10005];

int fa[10005];

inline void ds_init() {for(register int i = 1; i <= maxDate; i++) fa[i] = i;}

inline int ds_find(const int &u){
	if(fa[u] == u) return u;
	return fa[u] = ds_find(fa[u]);
}

inline void ds_union(const int &u, const int &v){
	int ancu = ds_find(u), ancv = ds_find(v);
	fa[ancu] = ancv;
}

int main(){
	while(scanf("%d", &n) != EOF){
		maxDate = 0;
		for(register int i = 1; i <= n; i++){
			scanf("%d%d", &prod[i].val, &prod[i].deadl);
			maxDate = max(maxDate, prod[i].deadl);
		}
		sort(prod + 1, prod + n + 1);
		ans = 0, ds_init();
		for(register int i = 1; i <= n; i++){
			int anc = ds_find(prod[i].deadl);
			if(anc) ans += prod[i].val, ds_union(anc, anc - 1);
		}
		printf("%d\n", ans);
	}
	return 0;
}

评论

还没有任何评论,你来说两句吧



常年不在线的QQ:
49750

不定期更新的GitHub:
https://github.com/Darkleafin


OPEN AT 2017.12.10

如遇到代码不能正常显示的情况,请刷新页面。
If the code cannot be displayed normally, please refresh the page.


发现一个优美的网站:
https://visualgo.net/en
















- Theme by Qzhai