timus[1090. In the Army Now]

看了一下題目的意思，求逆序數，求出逆序數最大的行號。

求逆序數是經典問題，想起當年剛學編程時，算法導論裏就有這個習題，如何將merge sort修改為可以計算逆序數的版本，於是寫了一下，得到了以下一份答案。

 1 #include <iostream>
 2 #include <algorithm>
 3 
 4 using namespace std;
 5 
 6 const int MAX_N = 10000;
 7 
 8 const int INF = MAX_N + 1;
 9 
10 int N, K;
11
 
 
12 int A[MAX_N];
13 
14 int aux[2][(MAX_N >> 1) + 1];
15 
16 int merge(int p, int mid, int q)
17 {
18 
19     int left_length = mid + 1 - p;
20     int right_length = q - mid;
21     for (int i = 0; i < left_length; ++i)
22     {
23         aux[0][i] = A[p + i];
24     }
25     aux[0][left_length] = INF;
 
26 
27     for (int i = 0; i < right_length; ++i)
28     {
29         aux[1][i] = A[mid + 1 + i];
30     }
31     aux[1][right_length] = INF;
32     int ans = 0;
33     int i = 0;
34     int j = 0;
35     for (int k = 0; k < left_length + right_length; ++k)
36     {
37         if (aux[0][i] < aux[1
 
][j])
38         {
39             A[p + k] = aux[0][i++];
40         }
41         else
42         {
43             A[p + k] = aux[1][j++];
44             ans += left_length - i;
45         }
46     }
47     return ans;
48 }
49 
50 int inversions(int p, int q)
51 {
52     int ans = 0;
53     if (p < q)
54     {
55         int mid = (p + q) / 2;
56         ans += inversions(p, mid);
57         ans += inversions(mid + 1, q);
58         ans += merge(p, mid, q);
59     }
60     return ans;
61 }
62 
63 int main()
64 {
65     cin >> N >> K;
66     int ans;
67     int max_inversions = -1;
68     for (int k = 0; k < K; ++k)
69     {
70         for (int i = 0; i < N; ++i)
71         {
72             cin >> A[i];
73         }
74         int x = inversions(0, N - 1);
75         if (x > max_inversions)
76         {
77             max_inversions = x;
78             ans = k + 1;
79         }
80     }
81     cout << ans << endl;
82 
83     return 0;
84 }

timus[1090. In the Army Now]