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

• 2018-01-04
• 0
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## 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:

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。。

Code #2 的时间复杂度为 O(Tnlogn)，这样才是正解。

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{

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{

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++){
}
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{

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++){
}
sort(prod + 1, prod + n + 1);
ans = 0, ds_init();
for(register int i = 1; i <= n; i++){
if(anc) ans += prod[i].val, ds_union(anc, anc - 1);
}
printf("%d\n", ans);
}
return 0;
}
```

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OPEN AT 2017.12.10

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