UVA1025-A Spy in the Metro(动态规划)

Posted 2021-01-09 npugen

Problem UVA1025-A Spy in the Metro

Accept: 713  Submit: 6160
Time Limit: 3000 mSec

Problem Description

 

 

Input

 

 

 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

 

题解：很明显的动态规划，dp[i][j]表示i时刻在j站点还需要的最短等待时间，总共就三种选择，状态转移方程很简单，边界一直是我写动态规划题比较头疼的地方，不过这个题还比较简单，t时刻在n站点自然是0，在其他站点就是INF（为了不会从这些状态转移过去）。

 

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxt = 200 + 10, maxn = 50 + 5;
 6 const int INF = 0x3f3f3f3f;
 7 
 8 int read() {
 9     int q = 0; char ch = ‘ ‘;
10     while (ch<‘0‘ || ch>‘9‘) ch = getchar();
11     while (‘0‘ <= ch && ch <= ‘9‘) {
12         q = q * 10 + ch - ‘0‘;
13         ch = getchar();
14     }
15     return q;
16 }
17 
18 int n, t, m1, m2;
19 int ti[maxn];
20 int dp[maxt][maxn];
21 bool have_train[maxt][maxn][2];
22 
23 void init() {
24     memset(have_train, false, sizeof(have_train));
25     for (int i = 1; i <= n - 1; i++) {
26         dp[t][i] = INF;
27     }
28     dp[t][n] = 0;
29 }
30 
31 int T = 1;
32 
33 int main()
34 {
35     //freopen("input.txt", "r", stdin);
36     while (scanf("%d", &n) && n) {
37         t = read();
38         init();
39         for (int i = 1; i < n; i++) {
40             ti[i] = read();
41         }
42         ti[n] = ti[0] = INF;
43         m1 = read();
44         int d;
45         for (int i = 0; i < m1; i++) {
46             d = read();
47             int cnt = 1;
48             for (int j = 1; j <= n; j++) {
49                 have_train[d][cnt][0] = true;
50                 //printf("d:%d cnt:%d
", d, cnt);
51                 cnt++;
52                 d += ti[j];
53             }
54         }
55         //printf("
");
56 
57         m2 = read();
58         for (int i = 0; i < m2; i++) {
59             d = read();
60             int cnt = n;
61             for (int j = n; j >= 1; j--) {
62                 have_train[d][cnt][1] = true;
63                 //printf("d:%d cnt:%d
", d, cnt);
64                 cnt--;
65                 d += ti[j - 1];
66             }
67         }
68 
69         for (int i = t - 1; i >= 0; i--) {
70             for (int j = 1; j <= n; j++) {
71                 dp[i][j] = dp[i + 1][j] + 1;
72                 if (i + ti[j] <= t && have_train[i][j][0]) {
73                     dp[i][j] = min(dp[i][j], dp[i + ti[j]][j + 1]);
74                 }
75 
76                 if (i + ti[j - 1] <= t && have_train[i][j][1]) {
77                     dp[i][j] = min(dp[i][j], dp[i + ti[j - 1]][j - 1]);
78                 }
79             }
80         }
81 
82         printf("Case Number %d: ", T++);
83         if (dp[0][1] >= INF) {
84             printf("impossible
");
85         }
86         else {
87             printf("%d
", dp[0][1]);
88         }
89     }
90     return 0;
91 }