POJ 2970 The lazy programmer（优先队列+贪心）

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Language:

The lazy programmer

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1566		Accepted: 386

Description

A new web-design studio, called SMART (Simply Masters of ART), employs two people. The first one is a web-designer and an executive director at the same time. The second one is a programmer. The director is so a nimble guy that the studio has already got N contracts for web site development. Each contract has a deadline d_i.

It is known that the programmer is lazy. Usually he does not work as fast as he could. Therefore, under normal conditions the programmer needs bi of time to perform the contract number i. Fortunately, the guy is very greedy for money. If the director pays him x_i dollars extra, he needs only (b_i − a_i x_i) of time to do his job. But this extra payment does not influent other contract. It means that each contract should be paid separately to be done faster. The programmer is so greedy that he can do his job almost instantly if the extra payment is (b_i ⁄ a_i) dollars for the contract number i.

The director has a difficult problem to solve. He needs to organize programmer’s job and, may be, assign extra payments for some of the contracts so that all contracts are performed in time. Obviously he wishes to minimize the sum of extra payments. Help the director!

Input

The first line of the input contains the number of contracts N (1 ≤ N ≤ 100 000, integer). Each of the next N lines describes one contract and contains integer numbers a_i, b_i, d_i (1 ≤ a_i, b_i ≤ 10 000; 1 ≤ d_i ≤ 1 000 000 000) separated by spaces.

Output

The output needs to contain a single real number S in the only line of file. S is the minimum sum of money which the director needs to pay extra so that the programmer could perform all contracts in time. The number must have two digits after the decimal point.

Sample Input

2
20 50 100
10 100 50

Sample Output

5.00

Source

Northeastern Europe 2004, Western Subregion

/*

题意:有n个任务要完成，每个任务有属性 a,b,d
	 分别代表  额外工资，完成时间，结束时间。
	 如果不给钱，那么完成时间为b,给x的额外工资，那么完成时间b变成 b-a*x

  求在所有任务在各自最后期限前完成所需要给的最少的钱

思路：先把任务按照结束之间排序，然后依次完成每个任务，如果这个任务无法完成
      那么在前面（包括这个）任务中找到a属性最大的任务（可以再给他钱的前提），
      给钱这个腾出时间来完成当前这个任务，而这个可以用优先队列维护


*/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 100005

struct stud{
  int a,b,d;
  double money;
  bool operator<(const stud b) const
  {
  	return a<b.a;
  }

}f[N];

int cmp(stud x,stud y)
{
	return x.d<y.d;
}

priority_queue<stud>q;

int n;

int main()
{
	int i,j;
	while(~scanf("%d",&n))
	{
		for(i=0;i<n;i++)
			{
				scanf("%d%d%d",&f[i].a,&f[i].b,&f[i].d);
			    f[i].money=0;   //已经给这个任务的钱
			}
	    sort(f,f+n,cmp);

	    while(!q.empty()) q.pop();
	    double ans,day;
	    ans=day=0;
	    stud cur;
	    for(i=0;i<n;i++)
		{
			q.push(f[i]);
			day+=f[i].b;
			while(day>f[i].d)
			{
				cur=q.top();
				q.pop();
				double temp=(double)(day-f[i].d)/cur.a; //完成这个任务需要给cur任务的钱
				if(temp+cur.money<(double)cur.b/cur.a)   //如果这个钱加上已经给的钱小于可以给他的钱
				{
					day-=temp*cur.a;
					cur.money+=temp;
					ans+=temp;
					q.push(cur);
					break;
				}
				else
				{
					temp=((double)cur.b/cur.a-cur.money);
					day-=temp*cur.a;
					ans+=temp;
				}
			}
		}

		printf("%.2f\n",ans);
	}
  return 0;
}

Language:

The lazy programmer