POJ 2970 The lazy programmer（優先隊列+貪心）

Language:Default 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 di. 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 xi dollars extra, he needs only (bi − ai xi) 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 (bi ⁄ ai) 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 ai, bi, di (1 ≤ ai, bi ≤ 10 000; 1 ≤ di ≤ 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

 

 

/*

題意:有n個任務要完畢，每個任務有屬性 a,b,d
	 分別表明  額外工資，完畢時間，結束時間。
	 假設不給錢。那麼完畢時間爲b,給x的額外工資，那麼完畢時間b變成 b-a*x

  求在所有任務在各自最後期限前完畢所需要給的最少的錢

思路：先把任務依照結束之間排序，而後依次完畢每個任務，假設這個任務沒法完畢
      那麼在前面（包含這個）任務中找到a屬性最大的任務（可以再給他錢的前提），
      給錢這個騰出時間來完畢當前這個任務，而這個可以用優先隊列維護


*/

#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;   //已經給這個任務的錢
			}
	    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; //完畢這個任務需要給cur任務的錢
				if(temp+cur.money<(double)cur.b/cur.a)   //假設這個錢加上已經給的錢小於可以給他的錢
				{
					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:Default 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 di. 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 xi dollars extra, he needs only (bi − ai xi) 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 (bi ⁄ ai) 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 ai, bi, di (1 ≤ ai, bi ≤ 10 000; 1 ≤ di ≤ 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