http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=35913
Secret agent Maria was sent to Algorithms City to carry out an especially dangerous mission. After
several thrilling events we find her in the first station of Algorithms City Metro, examining the time
table. The Algorithms City Metro consists of a single line with trains running both ways, so its time
table is not complicated.
Maria has an appointment with a local spy at the last station of Algorithms City Metro. Maria
knows that a powerful organization is after her. She also knows that while waiting at a station, she is
at great risk of being caught. To hide in a running train is much safer, so she decides to stay in running
trains as much as possible, even if this means traveling backward and forward. Maria needs to know
a schedule with minimal waiting time at the stations that gets her to the last station in time for her
appointment. 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. Trains
move in both directions: from the first station to the last station and from the last station back to the
first station. The time required for a train to travel between two consecutive stations is fixed since all
trains move at the same speed. Trains make a very short stop at each station, which you can ignore
for simplicity. Since she is a very fast agent, Maria can always change trains at a station even if the
trains 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 as
follows.
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 trains
between two consecutive stations: t1 represents the travel time between the first two stations, t2
the 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 first
station.
Line 5. M1 integers: d1, d2, . . . , dM1 (0 ≤ di ≤ 250 and di < di+1), representing the times at which
trains depart from the first station.
Line 6. The integer M2 (1 ≤ M2 ≤ 50), representing the number of trains departing from the N-th
station.
Line 7. M2 integers: e1, e2, . . . , eM2 (0 ≤ ei ≤ 250 and ei < ei+1) representing the times at which
trains 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 representing
the total waiting time in the stations for a best schedule, or the word ‘impossible’ in case Maria is
unable 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：紫书268页，
看着题解搞了一道wf题，爽！
dp[i][j]表示时刻i,在车站j,等待的最少时间
有3种方案:
等一分钟
往左搭车
往右搭车
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAX = ;
int t[MAX],n,T,M1,M2,d1[MAX],d2[MAX];
int has_train[MAX][MAX][],dp[MAX][MAX];
//has_train[i][j][0]表示i时刻在j车站往右走的车，has_train[i][j][1]表示往左行的车
int main()
{
int Case = ;
while(scanf("%d", &n) != EOF && n)
{
scanf("%d", &T);
for(int i = ; i < n; i++)
scanf("%d", &t[i]);
scanf("%d", &M1);
for(int i = ; i <= M1; i++)
scanf("%d", &d1[i]);
scanf("%d", &M2);
for(int j = ; j <= M2; j++)
scanf("%d", &d2[j]);
memset(has_train, , sizeof(has_train));
int sum = ;
//对has_train进行预处理
for(int i = ; i <= M1; i++)
{
sum = d1[i];
has_train[ d1[i] ][][] = ;
for(int j = ; j < n; j++)
{
sum += t[j];
has_train[ sum ][j + ][] = ;
}
}
for(int i = ; i <= M2; i++)
{
sum = d2[i];
has_train[ d2[i] ][n][] = ;
for(int j = n - ; j >= ; j--)
{
sum += t[j];
has_train[sum][j][] = ;
}
}
for(int i = ; i <= n-; i++)
dp[T][i] = INF;
dp[T][n] = ;
//这个第一层循环一定是从大往小循环，假设求i时刻j车站最少时间，在这点有三种情况，考虑往左走的车，那么选了往左走的车之后这一点的时间，前提是选了往左走的车辆，时间肯定是在i之后，由选了往左走的车后推出i,j；所以为什么要递减循环
for(int i = T - ; i >= ; i--)
{
for(int j = ; j <= n; j++)
{
dp[i][j] = dp[i + ][j] + ; //等待一分钟
if(j < n && has_train[i][j][] && i + t[j] <= T) //往右走，j必然要小于n，才能走，i+t[j]表示这一点的时间加上到下一点的时间要小于等于T，如果大于T没意义了，因为是T
{ dp[i][j] = min(dp[i][j], dp[ i + t[j] ][ j + ] );
}
if(j > && has_train[i][j][] && i + t[j - ] <= T)
{
dp[i][j] = min(dp[i][j], dp[ i + t[j - ] ][ j - ] );
}
}
}
printf("Case Number %d: ", ++Case);
if(dp[][] >= INF)
printf("impossible\n");
else
printf("%d\n",dp[][]);
} return ;
}