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Supermarket poj 1456 貪心+並查集優化

編輯：關於C++

Language: Supermarket Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9512 Accepted: 4096

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 Z喎?/kf/ware/vc/" target="_blank" class="keylink">vcHRpbWFsIHNlbGxpbmcgc2NoZWR1bGUgZm9yIGVhY2ggc2V0IG9mIHByb2R1Y3RzLiA8YnI+Cgo8cCBjbGFzcz0="pst">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

題意：超市有n個商品，每個商品有利潤p和保質期d，每天賣一種商品問怎麼賣才能使利潤最大，求出最大利潤。

思路：貪心，先按照商品利潤從大到小排序，選出利潤大的開始賣，賣的時間就在保質期d當天，若當天已經有商品在賣了，就從d往前推看哪一天可以賣該商品。

代碼：

//沒有並查集優化
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;

struct Node
{
    int p,d;
}node[maxn];

int n;
bool vis[maxn];

int cmp(Node a,Node b)
{
    return a.p>b.p;
}

int main()
{
    while (~scanf("%d",&n))
    {
        memset(vis,false,sizeof(vis));
        for (int i=0;i

並查集優化，在向前找哪一天能夠賣時使用並查集。
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;

struct Node
{
    int p,d;
}node[maxn];

int n;
int father[maxn];

int cmp(Node a,Node b)
{
    return a.p>b.p;
}

void init()
{
    for (int i=0;i<=maxn;i++)
        father[i]=i;
}

int find_father(int x)
{
    if (x!=father[x])
        father[x]=find_father(father[x]);
    return father[x];
}

int main()
{
    while (~scanf("%d",&n))
    {
        for (int i=0;i