1014 of greedy strategy

Problem O

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 71   Accepted Submission(s) : 13

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

22 10 1001020304050607080902 10 3103040

 

Sample Output

100 550 2

 题目要求：货船运送货车过河问题。给与货船最大装船量和货车的数目和到达时间，要求保证最小运送次数下的最短时间。

 思路：时间最短，即保证最后一辆货车最早到岸，而其前最大装船量内的货车不计入时间。进行递归调用后得到结果。m%n为0，直接n辆递归计算时间，否者先运前m%n辆然后n量递归求出时间。

细节：判断运完一次回岸后是否积累到最大装车量，用于计算时间。

#include<iostream>#include<iostream>#include<string.h>#include<set>#include<stdio.h>#include<vector>#include<algorithm>#include<numeric>#include<math.h>#include<string.h>#include<sstream>#include<stdio.h>#include<string>#include<cstdlib>#include<algorithm>#include<iostream>#include<map>#include<queue>#include<iomanip>#include<cstdio>using namespace std;int main(){int n, m, t,e,c,x,y,o,i;scanf("%d", &c);while (c--){scanf("%d%d%d", &n, &t, &m);x = m / n;o=m%n;if (o)x++;y=0;for(i=1;i<=m;i++)        {            scanf("%d",&e);            if(o)            {                if(i==o)y=e;                else if(i%n==o)            {             if(y+2*t>e)y+=2*t;             else y=e;            }            }            else            if(i==n){y=e;}            else if(i%n==0)            {                if(y+2*t>e)y+=2*t;             else y=e;            }         }y+=t;printf("%d %d\n", y, x);}}