poj 2336 Ferry Loading II ( 【贪心】 )

Ferry Loading II

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3704		Accepted: 1884

DescriptionBefore 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?
InputThe
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.
OutputFor
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.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
0
10
20
30
40
50
60
70
80
90
2 10 3
10
30
40

Sample Output

100 5
50 2

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

int et[1500];

int main()
{
int n, t, m;
int caset;
scanf("%d", &caset );
int i, j;
while(caset--)
{
scanf("%d %d %d", &n, &t, &m);
for(i=1; i<=m; i++)
{
scanf("%d", &et[i] );
}

//最早到达对岸的时间，取决于最后一辆车的被运送时间
//最优解是最后一辆车能够被尽早的运走

int dd, ff;
dd=m % n;
ff=m / n;
int ans=0, cnt=0;
if(dd!=0){
ans = et[dd];
ans = ans+t*2;
cnt++;
}
for(i=dd+n; i<=m; )
{
if( ans< et[i])
ans=et[i];
ans = ans+2*t;
i=i+n;
cnt++;
}
printf("%d %d\n", ans-t, cnt++ );

}
return 0;
}