BZOJ3427 Poi2013 Bytecomputer

可以YY一下嘛= =

最后一定是-1, -1, ..., -1, 0, 0, ... 0, 1, 1, ..., 1的一个数列

于是f[i][j]表示到了第i个数，新数列的第j项为-1 or 0 or 1的的最小代价

然后就没有然后了

 1 /**************************************************************
 2     Problem: 3427
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:808 ms
 7     Memory:16432 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13
14 using namespace std;
15 const int N = 1000005;
16 const int inf = 1e9;
17
18 int a[N], f[N][3];
19 int n, ans = inf;
20
21 inline int read() {
22     int x = 0, sgn = 1;
23     char ch = getchar();
24     while (ch < ‘0‘ || ‘9‘ < ch) {
25         if (ch == ‘-‘) sgn = -1;
26         ch = getchar();
27     }
28     while (‘0‘ <= ch && ch <= ‘9‘) {
29         x = x * 10 + ch - ‘0‘;
30         ch = getchar();
31     }
32     return sgn * x;
33 }
34
35 int main() {
36     int i, j, k;
37     int t, to;
38     n = read();
39     for (i = 1; i <= n; ++i)
40         a[i] = read();
41     memset(f, 127, sizeof(f));
42     f[1][a[1] + 1] = 0;
43     for (i = 1; i < n; ++i)
44         for (j = 0; j <= 2; ++j) if (f[i][j] < inf)
45             for (t = j - 1, k = 0; k <= 2; ++k) {
46                 to = a[i + 1] + k * t;
47                 if (to >= -1 && to <= 1 && to >= t)
48                     f[i + 1][to + 1] = min(f[i + 1][to + 1], f[i][j] + k);
49             }
50     for (i = 0; i <= 2; ++i)
51         ans = min(ans, f[n][i]);
52     if (ans == inf) puts("BRAK");
53     else printf("%d\n", ans);
54     return 0;
55 }