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1 | 1 | #include <bits/stdc++.h>
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2 |
| - |
3 | 2 | using namespace std;
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4 |
| - |
5 |
| -const int oo = 1000000000; |
6 |
| - |
7 |
| -const int N = 100005; |
8 |
| - |
9 |
| -int a[N], val[N][2], f[N][2], g[N][2]; |
10 |
| - |
11 |
| -void solve() { |
12 |
| - int n; |
13 |
| - scanf("%d", &n); |
14 |
| - |
15 |
| - for (int i = 1; i <= n; i++) { |
16 |
| - scanf("%d", a+i); |
17 |
| - } |
18 |
| - |
19 |
| - if (n == 1) { |
20 |
| - printf("0\n"); |
21 |
| - return; |
22 |
| - } |
23 |
| - |
24 |
| - for (int i = 1; i <= n; i++) { |
25 |
| - val[i][0] = a[i]; |
26 |
| - val[i][1] = a[n+1-i]; |
27 |
| - } |
28 |
| - |
29 |
| - int m = n/2; |
30 |
| - for (int i = 1; i <= m; i++) |
31 |
| - for (int k = 0; k < 2; k++) |
32 |
| - f[i][k] = g[i][k] = oo; |
33 |
| - f[1][0] = g[1][0] = 0; |
34 |
| - f[1][1] = g[1][1] = 1; |
35 |
| - for (int i = 1; i < m; i++) { |
36 |
| - for (int j = 0; j < 2; j++) { |
37 |
| - for (int k = 0; k < 2; k++) { |
38 |
| - int c1 = val[i][j]; |
39 |
| - int c2 = val[i+1][k]; |
40 |
| - if (((i & 1) && (c1 <= c2)) || (((i & 1) == 0) && c1 >= c2)) continue; |
41 |
| - int t = n+1-i; |
42 |
| - c1 = val[t][j]; |
43 |
| - c2 = val[t-1][k]; |
44 |
| - if (((t & 1) && (c1 <= c2)) || (((t & 1) == 0) && c1 >= c2)) continue; |
45 |
| - f[i+1][k] = min(f[i+1][k], f[i][j] + k); |
46 |
| - } |
47 |
| - for (int k = 0; k < 2; k++) { |
48 |
| - int c1 = val[i][j]; |
49 |
| - int c2 = val[i+1][k]; |
50 |
| - if (((i & 1) && (c1 >= c2)) || (((i & 1) == 0) && c1 <= c2)) continue; |
51 |
| - int t = n+1-i; |
52 |
| - c1 = val[t][j]; |
53 |
| - c2 = val[t-1][k]; |
54 |
| - if (((t & 1) && (c1 >= c2)) || (((t & 1) == 0) && c1 <= c2)) continue; |
55 |
| - g[i+1][k] = min(g[i+1][k], g[i][j] + k); |
56 |
| - } |
57 |
| - } |
58 |
| - } |
59 |
| - |
60 |
| - int ans = oo; |
61 |
| - |
62 |
| - if (n % 2 == 0) { |
63 |
| - for (int j = 0; j < 2; j++) { |
64 |
| - int c1 = val[m][j]; |
65 |
| - int c2 = val[m+1][j]; |
66 |
| - if (((m&1) && (c1 > c2)) || ((m&1) == 0 && (c1 < c2))) |
67 |
| - ans = min(ans, f[m][j]); |
68 |
| - if (((m&1) && (c1 < c2)) || ((m&1) == 0 && (c1 > c2))) |
69 |
| - ans = min(ans, g[m][j]); |
70 |
| - } |
71 |
| - } |
72 |
| - else { |
73 |
| - int c1 = a[m]; |
74 |
| - int c2 = a[m+2]; |
75 |
| - int b = a[m+1]; |
76 |
| - if ((((m+1)&1) && b > c1 && b > c2) || (((m+1)&1) == 0 && b < c1 && b < c2)) |
77 |
| - ans = min(ans, min(f[m][0], f[m][1])); |
78 |
| - if ((((m+1)&1) && b < c1 && b < c2) || (((m+1)&1) == 0 && b > c1 && b > c2)) |
79 |
| - ans = min(ans, min(g[m][0], g[m][1])); |
80 |
| - } |
81 |
| - |
82 |
| - if (ans >= oo) ans = -1; |
83 |
| - printf("%d\n", ans); |
| 3 | +int A[100009]; |
| 4 | +int D1[100009]; |
| 5 | +int D2[100009]; |
| 6 | +bool ok(int i) |
| 7 | +{ |
| 8 | + if (i % 2 == 1) |
| 9 | + return A[i] < A[i + 1]; |
| 10 | + else |
| 11 | + return A[i] > A[i + 1]; |
84 | 12 | }
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85 |
| - |
86 |
| -int main() { |
87 |
| - int ct; |
88 |
| - scanf("%d", &ct); |
89 |
| - |
90 |
| - while (ct--) solve(); |
| 13 | +int N; |
| 14 | +int ans() |
| 15 | +{ |
| 16 | + int i = (N + 1) / 2; |
| 17 | + int j = (N + 2) / 2; |
| 18 | + int id = 0; |
| 19 | + D1[id] = N + 2; |
| 20 | + D2[id] = N + 2; |
| 21 | + if (i == j) |
| 22 | + { |
| 23 | + D1[id] = 0; |
| 24 | + D2[id] = 0; |
| 25 | + } |
| 26 | + else |
| 27 | + { |
| 28 | + if (ok(i)) |
| 29 | + D1[id] = 0; |
| 30 | + swap(A[i], A[j]); |
| 31 | + if (ok(i)) |
| 32 | + D2[id] = 1; |
| 33 | + swap(A[i], A[j]); |
| 34 | + } |
| 35 | + while (true) |
| 36 | + { |
| 37 | + i--; |
| 38 | + j++; |
| 39 | + if (i < 1) |
| 40 | + break; |
| 41 | + id++; |
| 42 | + D1[id] = N + 2; |
| 43 | + D2[id] = N + 2; |
| 44 | + { |
| 45 | + if (ok(i) && ok(j - 1)) |
| 46 | + D1[id] = min(D1[id], D1[id - 1]); |
| 47 | + swap(A[i + 1], A[j - 1]); |
| 48 | + if (ok(i) && ok(j - 1)) |
| 49 | + D1[id] = min(D1[id], D2[id - 1]); |
| 50 | + swap(A[i + 1], A[j - 1]); |
| 51 | + } |
| 52 | + swap(A[i], A[j]); |
| 53 | + { |
| 54 | + if (ok(i) && ok(j - 1)) |
| 55 | + D2[id] = min(D2[id], 1 + D1[id - 1]); |
| 56 | + swap(A[i + 1], A[j - 1]); |
| 57 | + if (ok(i) && ok(j - 1)) |
| 58 | + D2[id] = min(D2[id], 1 + D2[id - 1]); |
| 59 | + swap(A[i + 1], A[j - 1]); |
| 60 | + } |
| 61 | + swap(A[i], A[j]); |
| 62 | + } |
| 63 | + return min(D1[id], D2[id]); |
| 64 | +} |
| 65 | +int main() |
| 66 | +{ |
| 67 | + int T; |
| 68 | + cin >> T; |
| 69 | + while (T--) |
| 70 | + { |
| 71 | + cin >> N; |
| 72 | + for (int i = 1; i <= N; i++) |
| 73 | + cin >> A[i]; |
| 74 | + int out = N + 2; |
| 75 | + out = min(out, ans()); |
| 76 | + for (int i = 1; i <= N; i++) |
| 77 | + A[i] *= -1; |
| 78 | + out = min(out, ans()); |
| 79 | + if (out > N) |
| 80 | + out = -1; |
| 81 | + cout << out << '\n'; |
| 82 | + } |
91 | 83 | }
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