Uva1025 A Spy in the Metro

Posted 2020-10-10 嘒彼小星

https://odzkskevi.qnssl.com/f42a2762ecb9e7541fde59742e2f52e7?v=1507521953

 

【题解】

DP。

dp[i][j]表示在i时间第j个车站最短等待时间

三种决策：

原地等1s

搭乘从左开往右的

搭乘从右开往左的

ok[i][j][0]表示第i车站时间j是否有从左开往右的

ok[i][j][1]表示第i车站时间j是否有从右开往左的

尼玛各种过不去，改的快跟std一样了还是不对，

最终把while(scanf("%d %d", &n, &T) != EOF && n && T)

改成 while(scanf("%d", &n) != EOF && n){read(T);}过了

我*****

浪费我两个多小时

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cmath>
 7 #define max(a, b) ((a) > (b) ? (a) : (b))
 8 #define min(a, b) ((a) < (b) ? (a) : (b))
 9 inline void swap(int &a, int &b)
10 {
11     int tmp = a;a = b;b = tmp;
12 }
13 inline void read(int &x)
14 {
15     x = 0;char ch = getchar(), c = ch;
16     while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();
17     while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
18     if(c == ‘-‘)x = -x;
19 }
20 
21 const int MAXN = 3000 + 5;
22 const int MAXT = 3000 + 10;
23 
24 int n,T,t[MAXN],m1,m2,ok[MAXN][MAXT][2];//0从左，1从右 
25 int dp[MAXN][MAXT];
26 
27 int tt;
28 
29 int main()
30 {
31     while(scanf("%", &n)!=EOF && n)
32     {
33                 read(T);
34         memset(ok, 0, sizeof(ok));
35         memset(t, 0, sizeof(t));
36         m1 = m2 = 0;
37         ++ tt;
38         for(register int i = 1;i < n;++ i)
39             read(t[i]);
40         read(m1);
41         for(register int i = 1;i <= m1;++ i)
42         {
43             int tmp;read(tmp);
44             for(register int j = 1;j <= n && tmp <= T;++ j)
45             {
46                 ok[j][tmp][0] = 1;
47                 tmp += t[j];
48             }
49         }
50         read(m2);
51         for(register int i = 1;i <= m2;++ i)
52         {
53             int tmp;read(tmp);
54             for(register int j = n;j >= 1 && tmp <= T;-- j)
55             {
56                 ok[j][tmp][1] = 1;
57                 tmp += t[j - 1];
58             }
59         }
60         memset(dp, 127, sizeof(dp));
61         dp[1][0] = 0;
62         for(register int i = 1;i <= T;++ i)
63             for(register int j = 1;j <= n;++ j)
64             {
65                 dp[j][i] = dp[j][i - 1] + 1;
66                 if(ok[j][i][0])
67                 {
68                     dp[j][i] = min(dp[j][i], dp[j - 1][i - t[j - 1]]);
69                 }
70                 if(ok[j][i][1])
71                 {
72                     dp[j][i] = min(dp[j][i], dp[j + 1][i - t[j]]);
73                 }
74             }
75         printf("Case Number %d: ", tt);
76         if(dp[n][T] >= 2139062143)printf("impossible\n");
77         else printf("%d\n", dp[n][T]);
78     }
79     return 0;
80 }    

Uva1025