Problem O
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?
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.
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.
10
3
#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);
}
}
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