ACM: uva 1467 - Installations

Installations

In the morning, service engineers in a telecom
company receive a list of jobs which they must serve today. They
install telephones, internet, ipTVs, etc and repair troubles with
established facilities. A client requires a deadline when the
requested job must be completed. But the engineers may not complete
some jobs within their deadlines because of job overload. For each
job, we consider, as a penalty of the engineer, the difference
between the deadline and the completion time. It measures how long
the job proceeds after its deadline. The problem is to find a
schedule minimizing the sum of the penalties of the jobs with the
two largest penalties.
A service engineer gets a list of
jobs Ji with a
serving
time si and a
deadline di. A
job Ji needs
time si, and if it is completed
at time Ci, then the penalty
of Ji is
defined to
be max{0, Ci - di}.
For convenience, we assume that the
time t when a job can be
served is 0
t < 
 and si and di are
given positive integers such that 0
< sidi. The goal is to find a schedule of jobs
minimizing the sum of the penalties of the jobs with the two
largest penalties.
For example, there are six
jobs Ji with
the
pair (si, di) of
the serving
time si and the
deadline di,i =
1,..., 6,
where (s1, d1)
= (1,
7), (s2, d2)
= (4,
7), (s3, d3)
= (2,
4), (s4, d4)
= (2,
15), (s5,d5) =
(3,
5), (s6, d6)
= (3, 8). Then Figure 1 represents a schedule which minimizes the
sum of the penalties of the jobs with the two largest penalties.
The sum of the two largest penalties of an optimal schedule is that
of the penalties
of J2 and J6,
namely 6 and 1, respectively, which is equal to 7 in this
example.



 
Figure 1. The optimal schedule of the example
Input 
Your program is to read from standard input. The
input consists of T test
cases. The number of test
cases T is given on the
first line of the input. The first line of each test case contains
an integern ( 1 n
500), the number of the given jobs. In the
next n lines of each test
case, the i-th line contains two integer
numbers si and di,
representing the serving time and the deadline of the
job Ji, respectively,
where 1 sidi
10, 000.
Output 
Your program is to write to standard output. Print
exactly one line for each test case. The line contains the sum of
the penalties of the jobs with the two largest penalties.
The following shows sample input and ouput for
three test cases.
Sample
Input 
3
6
1 7
4 7
2 4
2 15
3 5
3 8
7
2 17
2 11
3 4
3 20
1 20
4 7
5 14
10
2 5
2 9
5 10
3 11
3 4
4 21
1 7
2 9
2 11
2 23
Sample
Output 
7
0
14
 

题意: 有n个的任务要完成, 给出你它完成需要的时间和deadline, 惩罚值 = max(完成的时刻-deadline,
0);

     
需要你计算出最大的惩罚值和次大的惩罚值之和.(每个任务都有一个超时惩罚值)

 

解题思路:

     
1. 工作任务分配问题, 很容易想到将deadline小的任务先执行, 任务根据deadline排序.

     
2. 如果按照排序后从小到大执行一遍, 可以得出超时任务的最大和次大的惩罚值和它们的编号. 很容易

     
想到, 把超时的任务往前掉, 最后就是尝试把超时任务插入哪个位置使得最大和次大的惩罚值之和最小.

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

#define MAX 505

struct node

{

 int s, d;

 bool operator <(const node
&x) const

 {

  return d <
x.d;

 }

}a[MAX];

int n;

inline int max(int a, int b)

{

 return a > b ? a : b;

}

inline int min(int a, int b)

{

 return a < b ? a : b;

}

int solve()

{

 sort(a, a+n);

 int i, j;

 int c1 = 0, c2 = 0, pos = 0;

 int ans, cur = 0;

 for(i = 0; i < n; ++i)

 {

  cur += a[i].s;

  int temp = max(cur-a[i].d,
0);

  if(temp >
c1)

  {

   c2 =
c1;

   c1 =
temp;

   pos =
i;

  }

  else if(temp >
c2)

  {

   c2 =
temp;

   pos =
i;

  }

 }

 ans = c1+c2;

 for(i = 0; i < pos; ++i)

 {

  c1 = c2 = cur = 0;

  for(j = 0; j <
n; ++j)

  {

   if(i == j)
continue;

   cur +=
a[j].s;

   int temp =
max(cur-a[j].d, 0);

   if(temp
> c1)

   {

    c2
= c1;

    c1
= temp;

   }

   else if(temp
> c2) c2 = temp;

   if(j ==
pos)

   {

    cur
+= a[i].s;

    int
temp = max(cur-a[i].d, 0);

    if(temp
> c1)

    {

     c2
= c1;

     c1
= temp;

    }

    else
if(temp > c2) c2 = temp;

   }

  }

  ans = min(ans, c1+c2);

 }

 return ans;

}

int main()

{

// freopen("input.txt", "r", stdin);

 int caseNum;

 scanf("%d", &caseNum);

 while(caseNum--)

 {

  scanf("%d",
&n);

  for(int i = 0; i
< n; ++i)

   scanf("%d
%d", &a[i].s, &a[i].d);

  printf("%d\n",
solve());

 }

 return 0;

}