[UVA1025]城市里的间谍

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.

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.

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

题意：

某城市的地铁是线性的，有n（2≤n≤50）个车站,从左到右编号1-n。有M1辆列车从第1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1站出发,目的在时刻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。列车靠站停车时间忽略不计,且Mario身手敏捷,即时两辆方向不同的列车在同一时间靠站,Mario也能完成换乘。

输入第1行为n，第2行为T。

第3行有n-1个整数t1，t2，...，tn-1（1≤ti≤70），其中ti表示地铁从车站i到车站i+1的行驶时间（两个方向都一样）。

第4行为M1（1≤M1≤50），即从第1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＜di+1），即各列车的出发时间。

第6、7行描述从第n站出发向左开的列车，格式同第4、5行。

输出仅包含一行，即最少等待时间。无解输出impossible。

题解：

dp[i]: 在第i个车站, 时间为j时, 的总等待时间
三种状态:
dp[i][j]=dp[i][j-1]+1//在该车站停留1分钟
dp[i][j]=dp[i-1][j-t[i-1]]//车从左来(如果有)

dp[i][j]=dp[i+1][j-t[i]]//车从右来(如果有)

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int N=55;
const int M=205;
int n, T, m1, m2, dp[N][M], cas;
int t[N], rstart[N], lstart[N], rt[N], lt[N];
bool train[N][M][2];

void Pre() {
	rt[1]=lt[n]=0;
	for( int i=2; i<=n; i++ ) rt[i]=rt[i-1]+t[i-1];//向右从第1个站到第i个站的总时间
	for( int i=n-1; i>=1; i-- ) lt[i]=lt[i+1]+t[i];//向左从第n个站到第i个站的总时间
	
	memset( train, 0, sizeof train );
	for( int k=1; k<=m1; k++ )
		for( int i=2; i<=n; i++ )
			train[i][ rstart[k] + rt[i] ][0]=1;//train[i][j][0]: 在第i个站, 时间为j时, 是否有从左来的车
	for( int k=1; k<=m2; k++ )
		for( int i=n-1; i>=1; i-- )
			train[i][ lstart[k] + lt[i] ][1]=1;//train[i][j][1]: 在第i个站, 时间为j时, 是否有从右来的车
	
	memset( dp, INF, sizeof dp ); dp[1][0]=0;
}

void Getin() {
	scanf( "%d", &T );
	for( int i=1; i<n; i++ ) scanf( "%d", &t[i] );
	scanf( "%d", &m1 );
	for( int i=1; i<=m1; i++ ) scanf( "%d", &rstart[i] );
	scanf( "%d", &m2 );
	for( int i=1; i<=m2; i++ ) scanf( "%d", &lstart[i] );
	Pre();
}

int main() {
	while( ~scanf( "%d", &n ) ) {
		if( !n ) break;
		Getin();
		for( int j=1; j<=T; j++ )
			for( int i=1; i<=n; i++ ) {
				dp[i][j]=dp[i][j-1]+1;
				if( j>=t[i-1] && i>1 && train[i][j][0] )
					dp[i][j]=min( dp[i][j], dp[i-1][ j-t[i-1] ] );
				if( j>=t[i] && i<n && train[i][j][1] )
					dp[i][j]=min( dp[i][j], dp[i+1][ j-t[i] ] );
			}
		printf( "Case Number %d: ", ++cas );
		if( dp[n][T]>=dp[0][0] ) printf( "impossible\n" );
		else printf( "%d\n", dp[n][T] );
	}
	return 0;
}