Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3946		Accepted: 1985

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?

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.

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.

You may assume that 0 < n, t, m < 1440. The arrival times for each test case are in non-decreasing order.

2

2 10 10

0

10

20

30

40

50

60

70

80

90

2 10 3

10

30

40

100 5

50 2

题意：

给m辆车的到达岸边的时间,现在给你一个轮渡能运车的数量,和单程的时间,现在问把这些车运过去的最短时间是多少,在这个时间中的 最少运送次数是多少？

思路：

dp[i]表示运送前i个要用的时间,num[i]表示在dp[i]的时间内的最少次数;相邻的车在一块运,转移方程看代码吧;

AC代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

//#include <bits/stdc++.h>

#include <stack>

#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {

    char CH; bool F=false;

    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());

    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());

    F && (num=-num);

}

int stk[70], tp;

template<class T> inline void print(T p) {

    if(!p) { puts("0"); return; }

    while(p) stk[++ tp] = p%10, p/=10;

    while(tp) putchar(stk[tp--] + '0');

    putchar('\n');

}

const LL mod=1e9+7;

const double PI=acos(-1.0);

const int inf=1e9;

const int N=1e5+10;

const int maxn=2e3+14;

const double eps=1e-12;

int a[maxn],dp[maxn],num[maxn];

int main()

{

    int T;

    read(T);

    while(T--)

    {

        int n,m,t;

        read(n);read(t);read(m);

        For(i,1,m)read(a[i]);

        For(i,1,m)dp[i]=inf,num[i]=0;

        dp[0]=0;

        num[0]=0;

        For(i,1,m)

        {

            for(int j=max(0,i-n);j<i;j++)

            {

                if(j==0){dp[i]=a[i]+t,num[i]=1;continue;}

                if(dp[i]>max(dp[j]+t,a[i])+t)dp[i]=max(dp[j]+t,a[i])+t,num[i]=num[j]+1;

                else if(dp[i]==max(dp[j]+t,a[i])+t)num[i]=min(num[i],num[j]+1);

            }

        }

        cout<<dp[m]<<" "<<num[m]<<"\n";

    }

    return 0;

}