POJ-2336 Ferry Loading II（简单DP）

Ferry Loading II

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 3763 Accepted: 1919

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.

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

其实如果想到状态转移方程还是比较简单的，但是dp题目难的就是状态转移方程啊。我想了挺长时间，还是看了别人的博客。但是这道题目让我对dp题目又有了深刻的认识。我之前之所以想不出来是因为，我把状态弄混了，dp[i]表示第i辆车运送到对岸所花的时间，这个是

车的状态，而不是船的状态。所以所有的思考都应当围绕车子。这辆车被送到对岸，只有两种情况，要么是它不用等，船等着它，它一到就被带走。要么它等着穿，船一来它就跟船走。这就是车子的状态。

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

using namespace std;
#define MAX 1<<30
int dp[2000];
int bp[2000];
int a[2000];
int n,t,m;
int main()
{
    int c;
    scanf("%d",&c);
    while(c--)
    {
        scanf("%d%d%d",&n,&t,&m);
        for(int i=1;i<=m;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=m;i++)
            dp[i]=bp[i]=MAX;
        dp[0]=-t;
        dp[1]=a[1]+t;
        bp[1]=1;
        for(int i=2;i<=m;i++)
        {
            for(int j=max(0,i-n);j<i;j++)
            {
                dp[i]=min(dp[i],max(dp[j]+t,a[i])+t);
                bp[i]=min(bp[i],bp[j]+1);
            }
        }
        printf("%d %d\n",dp[m],bp[m]);
    }
    return 0;
}