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.
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
Analysis and understanding
"The Algorithms City Metro consists of a single line with trains running both ways" - two-way traffic, but it does not cycle
then address this issue, the wait time is the sum of the waiting time for each station -> recurrence of the station, with the only decision-making time and site-related
#include <bits/stdc++.h>
using namespace std;
int main(){
int n=1,T,m1,m2,i,j,fr_sum=0,bk_sum=0,sq=1;
while(~scanf("%d",&n),n){
scanf("%d",&T);
vector<int> t(n-1);
vector<vector<int>>fr_train(T+1,vector<int>(n));
vector<vector<int>>bk_train(T+1,vector<int>(n));
vector<vector<int>>dp(T+1,vector<int>(n));
for(auto &p:t)scanf("%d",&p);
scanf("%d",&m1);vector<int> fr(m1);
for(auto &p:fr){scanf("%d",&p);fr_sum+=p;}
scanf("%d",&m1);vector<int> bk(m1);
for(auto &p:bk){scanf("%d",&p);bk_sum+=p;}
for(auto &p:fr){
int x=0;
for(i=0;(x=p+i*fr_sum+i*bk_sum)<=T;++i){
fr_train[x][0]=1;
for(j=0;j<n-1;++j){
if((x+=t[j])<=T)fr_train[x][j+1]=1;
else break;
}
}
}
for(auto &p:bk){
int x=0;
for(i=0;(x=p+i*fr_sum+i*bk_sum)<=T;++i){
bk_train[x][n-1]=1;
for(j=n-1;j>0;--j){
if((x+=t[j-1])<=T)bk_train[x][j-1]=1;
else break;
}
}
}
dp[T][n-1]=0;
for(i=n-2;i>=0;--i)dp[T][i]=T+1;
for(i=T-1;i>=0;--i)
for(j=n-1;j>=0;--j){
dp[i][j]=dp[i+1][j]+1;
if(fr_train[i][j]==1&&i+t[j]<=T&&j+1<n){dp[i][j]=min(dp[i+t[j]][j+1],dp[i][j]);//printf("\nT%d:%d->%d",i,j,j+1);
}
if(bk_train[i][j]==1&&i+t[j-1]<=T&&j-1>=0){dp[i][j]=min(dp[i+t[j-1]][j-1],dp[i][j]);//printf("\nT%d:%d->%d",i,j,j-1);
}
}
printf("Case Number %d: ",sq);
if(dp[0][0]>T)printf("impossible\n");
else printf("%d\n",dp[0][0]);
sq++;
}
return 0;
},expression
Comma operator, the lowest priority, and above two equations it is coupled together, one by one calculation expression from left to right, and finally the entire expression is an expression. Such as: (8 + 3 + 5,6) is called comma expression, which expression solving process to 1, the expression 2, the entire value of the expression is the expression value of 2, such as: (3 + 5,6 + 8) value is 14; \ (a = (a =. 3 *. 5, a *. 4) \) values are 60, wherein \ ((a = 3 * 5 , a * 4) \) values are 60, the value of a comma expression has been in 15 years, the last 60 is assigned a comma expression is the value of a final 60.