Supermarket poj 1456 贪心+并查集优化

标签：supermarket   poj 1456   贪心+并查集优化   

Language: Default 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)=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

题意：超市有n个商品，每个商品有利润p和保质期d，每天卖一种商品问怎么卖才能使利润最大，求出最大利润。

思路：贪心，先按照商品利润从大到小排序，选出利润大的开始卖，卖的时间就在保质期d当天，若当天已经有商品在卖了，就从d往前推 看哪一天可以卖该商品。

代码：

//没有并查集优化
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#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<n;i++)
            scanf("%d%d",&node[i].p,&node[i].d);
        sort(node,node+n,cmp);
        int ans=0,f=-1;
        for (int i=0;i<n;i++)
        {
            int d=node[i].d;
            while (vis[d])
                d--;
            if (d<=0) continue;
            vis[d]=true;
            ans+=node[i].p;
        }
        printf("%d\n",ans);
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#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<n;i++)
            scanf("%d%d",&node[i].p,&node[i].d);
        init();
        sort(node,node+n,cmp);
        int ans=0;
        for (int i=0;i<n;i++)
        {
            int d=node[i].d;
            int dd=find_father(d);
            if (dd<=0) continue;
            father[dd]=dd-1;
            ans+=node[i].p;
        }
        printf("%d\n",ans);
    }
    return 0;
}

Supermarket poj 1456 贪心+并查集优化

标签：supermarket   poj 1456   贪心+并查集优化   

原文地址：http://blog.csdn.net/u014422052/article/details/43818109