UVA 1025 A Spy in the Metro （城市里的间谍（算法竞赛入门经典——例题9-1））（翻译，详解）

 

A Spy in the Metro

 

Secret agent Maria was sent to Algorithms City to carry out an especially dangerous mission. Afterseveral thrilling events we find her in the first station of Algorithms City Metro, examining the timetable. The Algorithms City Metro consists of a single line with trains running both ways, so its timetable is not complicated.Maria has an appointment with a local spy at the last station of Algorithms City Metro. Mariaknows that a powerful organization is after her. She also knows that while waiting at a station, she isat great risk of being caught. To hide in a running train is much safer, so she decides to stay in runningtrains as much as possible, even if this means traveling backward and forward. Maria needs to knowa schedule with minimal waiting time at the stations that gets her to the last station in time for herappointment. You must write a program that finds the total waiting time in a best schedule for Maria.The Algorithms City Metro system has N stations, consecutively numbered from 1 to N. Trainsmove in both directions: from the first station to the last station and from the last station back to thefirst station. The time required for a train to travel between two consecutive stations is fixed since alltrains move at the same speed. Trains make a very short stop at each station, which you can ignorefor simplicity. Since she is a very fast agent, Maria can always change trains at a station even if thetrains involved stop in that station at the same time.

Input

The input file contains several test cases. Each test case consists of seven lines with information asfollows.

Line 1. The integer N (2 ≤ N ≤ 50), which is the number of stations.

Line 2. The integer T (0 ≤ T ≤ 200), which is the time of the appointment.

Line 3. N − 1 integers: t1, t2, . . . , tN−1 (1 ≤ ti ≤ 20), representing the travel times for the trainsbetween two consecutive stations: t1 represents the travel time between the first two stations, t2the time between the second and the third station, and so on.

Line 4. The integer M1 (1 ≤ M1 ≤ 50), representing the number of trains departing from the firststation.

Line 5. M1 integers: d1, d2, . . . , dM1 (0 ≤ di ≤ 250 and di < di+1), representing the times at whichtrains depart from the first station.

Line 6. The integer M2 (1 ≤ M2 ≤ 50), representing the number of trains departing from the N-thstation.

Line 7. M2 integers: e1, e2, . . . , eM2 (0 ≤ ei ≤ 250 and ei < ei+1) representing the times at whichtrains depart from the N-th station.

The last case is followed by a line containing a single zero.

      某城市的地铁是线性的，有n（2<=n<=50）个车站，从左到右编号为1~n,有M1辆列车从第一站开始往右开，还有M2辆列车从第n站开始往左开。在时刻0，Mario从第一站出发，目的是在时刻T（0<=T<=200）会见车站n的一个间谍。在车站等车时容易被抓，所以她决定尽量躲在开动的火车上，让在车站等待的时间尽量短。列车到站停车时间忽略不计，且Mario身手敏捷，即使两辆方向不同的列车在同一时间靠站，Mario也能完成换乘。

                            

      

Output

For each test case, print a line containing the case number (starting with 1) and an integer representingthe total waiting time in the stations for a best schedule, or the word ‘impossible’ in case Maria isunable to make the appointment. Use the format of the sample output.

输入第1行为n，第2行为T，第3行有n-1个整数t1,t2,...t(n-1)（1<=ti<=70）,其中ti表示地铁从车站i到i+1的行驶时间（两个方向一样）。第4行为M1（1<=M1<=50），即从第一站出发向右开的列车数目。第5行包含M1个整数d1,d2,...dM1（0<=di<=250,di

Sample Input

4

55

5 10 15

4

0 5 10 20

4

0 5 10 15

4

18

1 2 3

5

0 3 6 10 12

6

0 3 5 7 12 15

2

30

20

1

20

7

1 3 5 7 11 13 17

0

Sample Output

Case Number 1: 5

Case Number 2: 0

Case Number 3: impossible

思路：

      时间是单向流逝的，是一个天然的“序”。影响到决策的只有当前时间和所处的车站，所以可以用dp[i][j]表示时刻i，你在车站j（编号为1~n），最少还需要等待多长时间。边界条件是dp[T][n]=0（已经见面），其他dp[T][j]（j!=n）为正无穷。有如下三种决策：

决策1：等一分钟

决策2：搭乘往右开的列车（如果有）

决策3：搭乘往左开的列车（如果有）

AC代码：

#include 
#include 
#include 
#define INF 0x3f3f3f3f
using namespace std;
int dp[205][55];//在时刻i，车站j最少还需等待多长时间能会面 
bool has_train[205][55][2];//has_train[i][j][0]（has_train[i][j][1]）在时刻i,车站j,是否有开往右（左）边的车 
int t[55];//从i到i+1站用的时间 
int n,T,M1,M2,d;//n个车站，在T时刻见面，M1列向左开的车，M2列向右开的车,发车时间 
int main()
{
    int count=1;
	while(~scanf("%d",&n))
	{
		if(n==0) return 0;
		memset(has_train,false,sizeof(has_train));
		scanf("%d",&T);
		for(int i=1;i<=n-1;i++)
		scanf("%d",&t[i]);
		scanf("%d",&M1);
		while(M1--)//处理向右开的列车，在i时刻j车站是否有向右开的列车 
		{
			scanf("%d",&d);//列车出发时刻 
			int i=d,j=1;
			while(j<=n)
			{
				has_train[i][j][0]=true; 
				i+=t[j];
				j++;
			}	
		} 
		scanf("%d",&M2);
		while(M2--)//处理向左开的列车，在i时刻j车站是否有向左开的列车
		{
			scanf("%d",&d);
			int i=d,j=n;
			while(j>=1)
			{
				has_train[i][j][1]=true;
				i+=t[j-1];
				j--;
			} 
		}
		for(int i=1;i<=n-1;i++)//在T时刻各个车站的情况初始 
		dp[T][i]=INF;
		//memset(dp,INF,sizeof(dp)); //也可设初值都是正无穷 
		dp[T][n]=0;//最终状态不用等待（在T时刻在车站n见面），这个不能忘了加 
		for(int i=T-1;i>=0;i--)//T-1个时刻 
		{
			for(int j=1;j<=n;j++)//n个车站
			{
				dp[i][j]=dp[i+1][j]+1;//等一分钟 
		        if(j1&&has_train[i][j][1]&&i+t[j-1]<=T)//（同理）这里是火车向左开的情况，t[j-1]是j-1站到j站（j站到j-1站）用的时间，也就是到上一站的时间 
		          dp[i][j]=min(dp[i][j],dp[i+t[j-1]][j-1]);//具备乘坐向左开列车的条件后，就可以判断是等一分钟还是花费t[j-1]时间乘坐向左开的列车到上一站，选最优情况 
			}	
		}
		printf("Case Number %d: ",count++); 
		if(dp[0][1]>=INF) printf("impossible\n");//没有方案 
		else printf("%d\n",dp[0][1]);//结果就是输出 在0时刻在第1站最少还需等待的时间 dp[0][1] 
	} 
	return 0;
}