Problem O

Problem Description

Before bridges
were common, ferries were used to transport cars across rivers.
River ferries, unlike their larger cousins, run on a guide line and
are powered by the river's current. Cars drive onto the ferry from
one end, the ferry crosses the river, and the cars exit from the
other end of the ferry.

There is a ferry across the river that can take n cars across the
river in t minutes and return in t minutes. m cars arrive at the
ferry terminal by a given schedule. What is the earliest time that
all the cars can be transported across the river? What is the
minimum number of trips that the operator must make to deliver all
cars by that time?

Input

The first line
of input contains c, the number of test cases. Each test case
begins with n, t, m. m lines follow, each giving the arrival time
for a car (in minutes since the beginning of the day). The operator
can run the ferry whenever he or she wishes, but can take only the
cars that have arrived up to that time.

Output

For each test
case, output a single line with two integers: the time, in minutes
since the beginning of the day, when the last car is delivered to
the other side of the river, and the minimum number of trips made
by the ferry to carry the cars within that time. < br><
br>You may assume that 0 < n, t, m < 1440. The arrival
times for each test case are in non-decreasing order.

Sample Input

2 2
10

10

10

20

30

40

50

60

70

80

90

2 10
3

10

30

40

Sample Output

100 5

50 2

题意：搬运车过河，但是并不是每一辆车都在一边等好的，每辆车到达岸边的时间有要求，一次最多搬运n辆车，需要时间t，并且回来也需要时间t，共有m辆车，求最少拌匀次数，且此时的用时。

解题思路：最少搬运次数不用想就是m%n?m/n+1:m/n;至于半时间如果m%n为零就就直接排好序之后每次搬n辆，要不的话就先把m%n搬走，让后面的车尽可能的少等；

感悟：做多了也没啥感悟了，最难的地方就是想贪心的条件；

代码（G++）

#include

#include

#define maxn 1444

using namespace std;

int main()

{

   
//freopen("in.txt", "r", stdin);

    int
c,t,n,m,wait_time[maxn],times=0,time=0,lost_car=0;

   
scanf("%d",&c);

    for(int
i=0;i

    {

       
time=times=0;

       
scanf("%d%d%d",&n,&t,&m);

       
//printf("n=%d t=%d m=%d\n",n,t,m);

       
for(int j=1;j<=m;j++)

           
scanf("%d",&wait_time[j]);

       
lost_car=m%n;

       
times=m%n?m/n+1:m/n;//最少运输的次数；

if(lost_car)

           
time=wait_time[lost_car]+t*2;

       
//printf("此时时间是%d\n",time);

       
for(int j=1;j<=m/n;j++)//总共运times次；

       
{

           
//printf("wait_time[j*n+lost_car]=%d\n",wait_time[j*n+lost_car]);

//printf("time=%d\n",time);

           
if(time

           
{

               
time+=2*t+(wait_time[j*n+lost_car]-time);

               
if(j==m/n)

                   
time-=t;//最后一次不用回去了

           
}

           
else//现在有车了

           
{

               
time+=2*t;

               
if(j==m/n)

               
time-=t;//最后一次不用回去了

           
}

           
//printf("此时时间是%d\n",time);

       
}

       
printf("%d %d\n",time,times);

    }

}