UVA1025-A Spy in the Metro(动态规划)

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）。

 #include <bits/stdc++.h>

 using namespace std;

 const int maxt =  + , maxn =  + ;
 const int INF = 0x3f3f3f3f;

 int read() {
     int q = ; char ch = ' ';
     while (ch<'' || ch>'') ch = getchar();
     while ('' <= ch && ch <= '') {
         q = q *  + ch - '';
         ch = getchar();
     }
     return q;
 }

 int n, t, m1, m2;
 int ti[maxn];
 int dp[maxt][maxn];
 bool have_train[maxt][maxn][];

 void init() {
     memset(have_train, false, sizeof(have_train));
     for (int i = ; i <= n - ; i++) {
         dp[t][i] = INF;
     }
     dp[t][n] = ;
 }

 int T = ;

 int main()
 {
     //freopen("input.txt", "r", stdin);
     while (scanf("%d", &n) && n) {
         t = read();
         init();
         for (int i = ; i < n; i++) {
             ti[i] = read();
         }
         ti[n] = ti[] = INF;
         m1 = read();
         int d;
         for (int i = ; i < m1; i++) {
             d = read();
             int cnt = ;
             for (int j = ; j <= n; j++) {
                 have_train[d][cnt][] = true;
                 //printf("d:%d cnt:%d\n", d, cnt);
                 cnt++;
                 d += ti[j];
             }
         }
         //printf("\n");

         m2 = read();
         for (int i = ; i < m2; i++) {
             d = read();
             int cnt = n;
             for (int j = n; j >= ; j--) {
                 have_train[d][cnt][] = true;
                 //printf("d:%d cnt:%d\n", d, cnt);
                 cnt--;
                 d += ti[j - ];
             }
         }

         for (int i = t - ; i >= ; i--) {
             for (int j = ; j <= n; j++) {
                 dp[i][j] = dp[i + ][j] + ;
                 if (i + ti[j] <= t && have_train[i][j][]) {
                     dp[i][j] = min(dp[i][j], dp[i + ti[j]][j + ]);
                 }

                 if (i + ti[j - ] <= t && have_train[i][j][]) {
                     dp[i][j] = min(dp[i][j], dp[i + ti[j - ]][j - ]);
                 }
             }
         }

         printf("Case Number %d: ", T++);
         if (dp[][] >= INF) {
             printf("impossible\n");
         }
         else {
             printf("%d\n", dp[][]);
         }
     }
     return ;
 }