zoukankan      html  css  js  c++  java
  • uva 1025 A Spy in the Metro 解题报告

    Time Limit: 3000MS     64bit IO Format: %lld & %llu

     Status uDebug

      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

    ——————————————————我是分割线————————————————————

    DP水题。

      1 #include<iostream>
      2 #include<cstdio>
      3 #include<cstring>
      4 #include<cmath>
      5 #include<algorithm>
      6 #include<queue>
      7 #include<cstdlib>
      8 #include<iomanip>
      9 #include<cassert>
     10 #include<climits>
     11 #include<functional>
     12 #include<bitset>
     13 #include<vector>
     14 #include<list>
     15 #define maxn 51
     16 #define F(i,j,k) for(int i=j;i<=k;i++)
     17 #define M(a,b) memset(a,b,sizeof(a))
     18 #define FF(i,j,k) for(int i=j;i>=k;i--)
     19 #define inf 0x7fffffff
     20 #define maxm 1001
     21 #define mod 998244353
     22 //#define LOCAL
     23 using namespace std;
     24 int read(){
     25     int x=0,f=1;char ch=getchar();
     26     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
     27     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
     28     return x*f;
     29 }
     30 int dp[220][55];
     31 int t_use[55];
     32 vector<int> go[220][55];
     33 int main() 
     34 {
     35     int n,T,icase=1,times;
     36     while ((cin>>n)&&n) {
     37         cin>>T;
     38         int m1,m2;
     39         for (int i=1;i<=n-1;++i) cin>>t_use[i];
     40         for (int i=0;i<=T;++i) 
     41             for (int j=1;j<=n;++j)
     42                 go[i][j].clear();
     43         cin>>m1;
     44         for (int i=1;i<=m1;++i) {
     45             cin>>times;
     46             int t=times;
     47             if(t>T) continue;
     48             go[t][1].push_back(i);
     49             for (int j=1;j<=n-1;++j) {
     50                 t+=t_use[j];
     51                 if (t>T) break;
     52                 go[t][j+1].push_back(i);
     53             }
     54         }
     55         cin>>m2;
     56         for (int i=1;i<=m2;++i) {
     57             cin>>times;
     58             int t=times;
     59             if (t>T) continue;
     60             go[t][n].push_back(i+m1);
     61             for (int j=n-1;j>=1;--j) {
     62                 t+=t_use[j];
     63                 if(t>T) break;
     64                 go[t][j].push_back(i+m1);
     65             }
     66         }
     67         for (int i=0;i<=T;++i)
     68             for (int j=1;j<=n;++j)
     69                 dp[i][j]=inf;
     70         dp[0][1]=0;
     71         for (int i=0;i<T;++i) {
     72             for (int j=1;j<=n;++j) {
     73                 if (dp[i][j]==inf) continue;
     74                 int size=go[i][j].size();
     75                 dp[i+1][j]=min(dp[i+1][j],dp[i][j]+1);
     76                 for (int k=0;k<size;++k) {
     77                     int u=go[i][j][k];
     78                     if (u<=m1&&j<n) {
     79                         if (i+t_use[j]<=T) {
     80                             dp[i+t_use[j]][j+1]=min(dp[i+t_use[j]][j+1],dp[i][j]);
     81                         }
     82                     }
     83                     else if (j>1&&u>m1) {
     84                         if (i+t_use[j-1]<=T) {
     85                             dp[i+t_use[j-1]][j-1]=min(dp[i+t_use[j-1]][j-1], dp[i][j]);
     86                         }
     87                     }
     88                 }
     89             }
     90         }
     91         int ans=inf;
     92         for (int i=0;i<=T;++i) {
     93             if (dp[i][n]==inf) continue;
     94             ans=min(ans,dp[i][n]+T-i);
     95         }
     96         if (ans==inf) {
     97             cout<<"Case Number "<<icase++<<": impossible"<<endl;
     98         }
     99         else {
    100             cout<<"Case Number "<<icase++<<": "<<ans<<endl;
    101         }
    102     }
    103     return 0;
    104 }
    105 /*
    106 4
    107 55
    108 5 10 15
    109 4
    110 0 5 10 20
    111 4
    112 0 5 10 15
    113 4
    114 18
    115 1 2 3
    116 5
    117 0 3 6 10 12
    118 6
    119 0 3 5 7 12 15
    120 2
    121 30
    122 20
    123 1
    124 20
    125 7
    126 1 3 5 7 11 13 17
    127 0
    128 */
    uva 1025

     

  • 相关阅读:
    10年测试专家深度解读接口测试
    测试技术大牛谈成长经历:一个好的软件测试工程师应该做到这些!
    一位测试老鸟的工作经验分享
    又一名程序员倒下,网友:我们只是新时代农民工
    软件测试工程师这样面试,拿到offer的几率是80%
    App测试流程及测试点(个人整理版)
    自动化测试是什么?
    软件测试工程师的职业技能分析
    月薪15k的测试员需要学习什么技术?
    面向对象
  • 原文地址:https://www.cnblogs.com/SBSOI/p/5842266.html
Copyright © 2011-2022 走看看