动态规划——地铁里的间谍

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：

共有n个车站，现在你要在t时刻到n站和一个间谍见面，告诉你每两个相邻车站之间的行驶时间(无论是从左往右还是从右往左，时间不变)，再告诉有m1辆车会从第1站出发向第n站行驶，并给出这m1辆车的出发时间。然后再告诉有m2辆车会从第n站出发向第1站行驶，并给出这m2辆车的出发时间，现在要求出你所要等待的最小时间。在一辆行驶的车上所花的时间不算在等待时间里面，忽略上下车和列车停靠的时间。

思路：

由于影响决策的只有当前的时刻和当前所处的站，于是用dp[i][j]表示在i时刻时我在j站所需等待的最小时间。

那么有三种决策

1.在原地等1分钟

2.如果此时有向右开的车，就上车

3.如果此时有向左开的车，就上车

事先将dp数组置为INF，之后判断dp[0][1]的值的情况看是否有解即可

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <stack>#define INF 0x3f3f3f3f#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int maxn=55;const int maxt=220;int n ,t;bool train[maxt][maxn][2];int timetable[maxn];int dp[maxt][maxn];void solve(){    for(int i=1; i<n; i++)        dp[t][i]=INF;    dp[t][n]=0;    for(int i=t-1; i>=0; i--)    {        for(int j=1; j<=n; j++)        {            dp[i][j]=dp[i+1][j]+1;   //决策1            if((i+timetable[j]<=t)&&j<n&&train[i][j][0])   //决策2                dp[i][j]=min(dp[i][j], dp[i+timetable[j]][j+1]);            if((i+timetable[j-1])<=t&&j>1&&train[i][j][1])  //决策3                dp[i][j]=min(dp[i][j], dp[i+timetable[j-1]][j-1]);        }    }}int main(){    int cas=0;    while(~scanf("%d", &n) &&n)    {        scanf("%d", &t);        memset(timetable, 0, sizeof(timetable));        memset(train, 0, sizeof(train));        for(int i=1; i<n; i++)            scanf("%d", &timetable[i]);   //存放相邻两个站之间的行驶时间        int m1, m2;        scanf("%d", &m1);        for(int i=0; i<m1; i++)   //计算每个站有车的时刻        {            int x;            scanf("%d", &x);            train[x][1][0]=true;            for(int j=1; j<n; j++)                if(x+timetable[j]<=t)    //不加会RE                    train[x+=timetable[j]][j+1][0]=true;        }        scanf("%d", &m2);        for(int i=0; i<m2; i++)   //同上        {            int x;            scanf("%d", &x);            train[x][n][1]=true;            for(int j=n-1; j>=1; j--)                if(x+timetable[j]<=t)                    train[x+=timetable[j]][j][1]=true;        }        solve();        if(dp[0][1]>=INF)            printf("Case Number %d: impossible\n", ++cas);        else            printf("Case Number %d: %d\n", ++cas, dp[0][1]);    }    return 0;}