POJ 1456 Supermarket 区间问题并查集||贪心

F - Supermarket

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1456

Appoint description: 
System Crawler  (2015-11-30)

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
1 本题完全可以贪心解决，按照价值从大到小进行排列，每次取最大的价值，从他结束的时间开始向前暴力，如果有时间可以使用，没被vis数组标记，就可以再次位置进行
2  上述是贪心的思想，下面可以使用并查集进行优化，还是按照价值问题进行排列，每次回溯，找前面有一个点father【i】==i，那么此点可以使用，就可以停止进行，此外此点自己--
表示该时间也被使用，后面如果回溯到该点的时候不可以停止下来，仍然需要向前回溯，如果一直回溯到0，那么就是没有位置可以进行，那么这次的价值也不会算到总和里面
下面分别贴上两种代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
struct node{
    int t;
    int price;
}que[maxn];
bool vis[maxn];
bool cmp(const struct node t1,const struct node t2){
    if(t1.price==t2.price)
        return t1.t>t2.t;
    return t1.price>t2.price;
}

int main(){
    int n;

    while(scanf("%d",&n)!=EOF){
        memset(vis,false,sizeof(vis));

        for(int i=;i<n;i++){
           scanf("%d%d",&que[i].price,&que[i].t);
        }
        sort(que,que+n,cmp);
        int sum=;

        for(int i=;i<n;i++){
            int tmp_t=que[i].t;
            for(int j=tmp_t;j>=;j--){
               if(!vis[j]){
                  vis[j]=true;
                  sum+=que[i].price;
                  break;
               }
            }
        }
        printf("%d\n",sum);
    }
    return ;
}

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
int father[maxn];
struct node{
    int value;
    int t;
}que[maxn];
void init(){
   for(int i=;i<maxn;i++)
       father[i]=i;
}

bool operator <(const node& t1,const node& t2){
   if(t1.value!=t2.value)
       return t1.value>t2.value;
   return t1.t>t2.t;
}
int get_father(int x){
    if(father[x]!=x)
        father[x]=get_father(father[x]);
    return father[x];
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
         init();
         for(int i=;i<n;i++){
            scanf("%d%d",&que[i].value,&que[i].t);
         }
         sort(que,que+n);
         int ans=;
         for(int i=;i<n;i++){
           int tmp=que[i].t;
           int x=get_father(tmp);
           if(x>){
                 father[x]--;
                 ans+=que[i].value;
           }
         }
         printf("%d\n",ans);
    }
   return ;
}