题目
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 structions
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
分析与理解
"The Algorithms City Metro consists of a single line with trains running both ways"--双向通行，但并不循环
然后针对这个问题，等待时间是各站等待时间之和-->对站递推,决策只与时间和站点有关
代码
#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]);
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("Case Number %d: ",sq);
if(dp[0][0]>T)printf("impossible
");
else printf("%d
",dp[0][0]);
sq++;
}
return 0;
}
,表达式
逗号运算符，优先级别最低，它将两个及其以上的式子联接起来，从左往右逐个计算表达式，整个表达式的值为最后一个表达式的值。如：（3+5,6+8）称为逗号表达式，其求解过程先表达式1，后表达式2，整个表达式值是表达式2的值，如：（3+5，6+8）的值是14；(a=(a=3*5,a*4))的值是60，其中(（a=3*5,a*4）)的值是60， a的值在逗号表达式里一直是15，最后被逗号表达式赋值为60，a的值最终为60。