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 ", ans); 54 return 0; 55 }