BZOJ 4302 Buildings 解题报告

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int Fx[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
 6 int n, m, x, y;
 7 
 8 inline int Calc(int tx, int ty, int k)
 9 {
10     int Min = min(n, m);
11     for (int i = 0; i < 4; i ++)
12     {
13         if (k == i) continue ;
14         if (i == 0) Min = min(Min, n + 1 - tx);
15         if (i == 1) Min = min(Min, tx);
16         if (i == 2) Min = min(Min, m + 1 - ty);
17         if (i == 3) Min = min(Min, ty);
18     }
19     return Min;
20 }
21 
22 inline int Solve()
23 {
24     if (n == m && (n & 1) && x == y && (x * 2 - 1 == n))
25         return n - 1 >> 1;
26     int ans = min(n, m) + 1 >> 1;
27     for (int k = 0; k < 4; k ++)
28     {
29         int tx = x + Fx[k][0], ty = y + Fx[k][1];
30         if (tx && ty && tx <= n && ty <= m)
31             ans = max(ans, Calc(tx, ty, k ^ 1));
32     }
33     return ans;
34 }
35 
36 int main()
37 {
38     while (scanf("%d%d%d%d", &n, &m, &x, &y) == 4)
39         printf("%d\n", Solve());
40     
41     return 0;
42 }