poj 1456(贪心并查集)

Posted 2020-11-09 flyer_duck

Supermarket

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15416		Accepted: 6941

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)=Σ_x∈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.

 

水题。可以直接用贪心，但是为了深刻的理解并查集，用并查集优化。可提高速度

第一种，直接贪心

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn=10010;
 6 struct node{
 7     int p,d;
 8 }a[maxn];
 9 
10 bool cmp(node x,node y)
11 {
12     return x.p>y.p;
13 }
14 bool flag[maxn];
15 
16 int main()
17 {
18     int n;
19     while( ~scanf("%d",&n)){
20         memset( a, 0, sizeof a);
21         for(int i=1;i<=n;i++){
22             scanf("%d%d",&a[i].p,&a[i].d);
23         }
24         sort( a+1, a+1+n, cmp);
25 //        for(int i=1;i<=n;i++)
26 //            printf("%d %d\n",a[i].p, a[i].d);
27         memset( flag, 0, sizeof flag);
28         int ans=0;
29         for(int i=1;i<=n;i++){
30             if(!flag[a[i].d]){
31                 flag[a[i].d]=true;
32                 ans+=a[i].p;
33             } else{
34                 for(int j=a[i].d;j>0;j--){
35                     if(!flag[j]){
36                         flag[j]=true;
37                         ans+=a[i].p;
38                         break;
39                     }
40                 }
41             }
42         }
43         printf("%d\n",ans);
44     }
45     return 0;
46 }

用并查集优化

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn=10010;
 6 int fa[maxn];
 7 struct node{
 8     int p,d;
 9 }a[maxn];
10 
11 bool cmp(node x,node y)
12 {
13     return x.p>y.p;
14 }
15 int find_fa(int x)
16 {
17     return fa[x]==x?x:(fa[x]=find_fa(fa[x]));
18 }
19 
20 int main()
21 {
22     int n;
23     while( ~scanf("%d",&n)){
24         for(int i=0;i<maxn;i++)
25             fa[i]=i;
26         for(int i=1;i<=n;i++){
27             scanf("%d%d",&a[i].p,&a[i].d);
28         }
29         sort( a+1, a+1+n, cmp);
30 
31         int ans=0;
32         for(int i=1;i<=n;i++){
33             int r=find_fa(a[i].d);
34             if(r>0){
35                 fa[r]=r-1;
36                 ans+=a[i].p;
37             }
38         }
39         printf("%d\n",ans);
40     }
41     return 0;
42 }