﻿POJ 1456 Supermarket(贪心|并查集)-Ocrosoft

# POJ 1456 Supermarket(贪心|并查集)

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 11462 Accepted: 5058

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.

Solution

```#include <cstdio>
#include <iostream>
#include <stack>
#include <string>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=10010;
struct UF//并查集，包含递归和非递归
{
int pre[MAXN];
int find(int x)
{
if(pre[x]==x)return x;//是根节点
pre[x]=find(pre[x]);//（路径压缩）不是根节点，把父节点改为根节点
return pre[x];//并返回父节点
}
int ufind(int x)
{
int r=x;
while(pre[r]!=r)r=pre[r];
int i=x,j;
while(i!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void uni(int x,int y)
{
pre[find(x)]=find(y);//x的根节点记为y的根节点
}
void uuni(int x,int y)
{
pre[ufind(x)]=ufind(y);//x的根节点记为y的根节点
}
};
struct node
{
int v;//value
} t;
vector<node> v;
int cmp(node a,node b)
{
return a.v>b.v;
}
bool vis[MAXN];
int main()
{
int n;
while(cin>>n)
{
memset(vis,0,sizeof(vis));
int maxdate=0;
v.clear();
for(int i=0; i<n; i++)
{
cin>>t.v>>t.t;
v.push_back(t);
maxdate=max(maxdate,t.t);
}
sort(v.begin(),v.end(),cmp);
int ans=0;
for(int i=0; i<n; i++)
{
if(!vis[v[i].t])
{
ans+=v[i].v;
vis[v[i].t]=1;
}
else
{
for(int j=v[i].t-1; j>=1; j--)
{
if(!vis[j])
{
ans+=v[i].v;
vis[j]=1;
break;
}
}
}
}
cout<<ans<<endl;
}
return 0;
}
```

```#include <cstdio>
#include <iostream>
#include <stack>
#include <string>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=10010;
struct UF//并查集，包含递归和非递归
{
int pre[MAXN];
int find(int x)
{
if(pre[x]==x)return x;//是根节点
pre[x]=find(pre[x]);//（路径压缩）不是根节点，把父节点改为根节点
return pre[x];//并返回父节点
}
int ufind(int x)
{
int r=x;
while(pre[r]!=r)r=pre[r];
int i=x,j;
while(i!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
void uni(int x,int y)
{
pre[find(x)]=find(y);//x的根节点记为y的根节点
}
void uuni(int x,int y)
{
pre[ufind(x)]=ufind(y);//x的根节点记为y的根节点
}
void ini()
{
for(int i=0;i<MAXN;i++)
pre[i]=i;
}
}uf;
struct node
{
int v;//value
} t;
vector<node> v;
int cmp(node a,node b)
{
return a.v>b.v;
}
bool vis[MAXN];
int main()
{
int n;
while(cin>>n)
{
memset(vis,0,sizeof(vis));
uf.ini();
int maxdate=0;
v.clear();
for(int i=0; i<n; i++)
{
cin>>t.v>>t.t;
v.push_back(t);
maxdate=max(maxdate,t.t);
}
int ans=0;
sort(v.begin(),v.end(),cmp);
for(int i=0;i<n;i++)
{
int r=uf.find(v[i].t);
if(r)//不是0表示前面还有某些天是空的
{
uf.pre[r]=r-1;//指到前一天，表示这一天已经被使用了
ans+=v[i].v;
}
}
cout<<ans<<endl;
}
return 0;
}
```