NYOJ 208 并查集的强大应用

        做了这道题，体会到了并查集的强大啊，，，同时也体会到了对并查集的了解和应用的欠缺。。。。。还是太弱啊。。。。。。。。完全想不到的，竟然可以用并查集做，，，神奇。。。。。。。题目：

Supermarket

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

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. 

输入

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.

输出

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.

样例输入

4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8 2
   5 20  50 10

样例输出

80
185

ac代码：

 
#include <iostream>
#include <string.h>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N=10010;
int father[N];
struct good{
  int value;
  int day;
}aa[N];
bool cmp(good a,good b){
	return a.value<b.value;
}
int find(int x){
  if(x!=father[x])
	  father[x]=find(father[x]);
  return father[x];
}
int main(){
	//freopen("2.txt","r",stdin);
  int n;
  while(~scanf("%d",&n)){
    for(int i=0;i<N;++i)
		father[i]=i;
	int sum=0;
	for(int i=0;i<n;++i){
		scanf("%d%d",&aa[i].value,&aa[i].day);
	}
	sort(aa,aa+n,cmp);
	for(int i=n-1;i>=0;--i){
		int b=find(father[aa[i].day]);
		//printf("b===%d\n",b);
		if(b>0){
			sum+=aa[i].value;
			father[b]=b-1;
			//printf("father[%d]==%d\n",aa[i].day,b-1);
		}
	}
	printf("%d\n",sum);
  }
  return 0;
}        

 

另一种方法：

 
#include <iostream>
#include <string.h>
#include <algorithm>
#include <cstdio>
using namespace std;
const int N=10010;
struct good{
   int value;
   int day;
}aa[N];
int flag[N];
bool cmp(good a,good b){
	return a.value<b.value;
}
int main(){
	//freopen("2.txt","r",stdin);
	int n;
	while(~scanf("%d",&n)){
		memset(flag,0,sizeof(flag));
	  for(int i=0;i<n;++i)
		  scanf("%d%d",&aa[i].value,&aa[i].day);
	  sort(aa,aa+n,cmp);
	  int sum=0;
	  for(int i=n-1;i>=0;--i){
		  for(int j=aa[i].day;j>=1;--j){
			  if(!flag[j]){
				  sum+=aa[i].value;
				  flag[j]=1;
			    break;
			  }
		  }
	  }
	  printf("%d\n",sum);
	}
  return 0;
}