LA 3695 Distant Galaxy

 1 //#define LOCAL
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm> 
 5 using namespace std;
 6 
 7 const int maxn = 100 + 10;
 8 int n, m, y[maxn], on[maxn], on2[maxn], left[maxn];
 9 
10 struct Point
11 {
12     int x, y;
13     bool operator < (const Point& rhs) const
14     {
15         return x < rhs.x;
16     }
17 }p[maxn];
18 
19 int solve()
20 {
21     sort(p, p + n);
22     sort(y, y + n);
23     m = unique(y, y + n) - y;    //m为不同y坐标的个数
24     if(m <= 2)
25         return n;
26 
27     int ans = 0;
28     for(int a = 0; a < m; ++a)
29         for(int b = a + 1; b < m; ++b)
30         {
31             int ymin = y[a], ymax = y[b];
32 
33             //计算left, on, on2
34             int k = 0;    //k记录竖线的条数
35             for(int i = 0; i < n; ++i)
36             {
37                 if(i == 0 || p[i].x != p[i-1].x)
38                 {    //这是一条新的竖线
39                     ++k;
40                     on[k] = on2[k] = 0;
41                     left[k] = k == 0 ? 0 : left[k-1] + on2[k-1] - on[k-1];
42                 }
43                 if(p[i].y > ymin && p[i].y < ymax)
44                     ++on[k];
45                 if(p[i].y >= ymin && p[i].y <= ymax)
46                     ++on2[k];
47             }
48             if(k <= 2)
49                 return n;
50 
51             int M = 0;
52             for(int j = 1; j <= k; ++j)
53             {
54                 ans = max(ans, left[j] + on2[j] + M);
55                 M = max(M, on[j] - left[j]);
56             }
57         }
58 
59     return ans;
60 }
61 
62 int main(void)
63 {
64     #ifdef LOCAL
65         freopen("3695in.txt", "r", stdin);
66     #endif
67 
68     int kase = 0;
69     while(scanf("%d", &n) == 1 && n)
70     {
71         for(int i = 0; i < n; ++i)
72         {
73             scanf("%d%d", &p[i].x, &p[i].y);
74             y[i] = p[i].y;
75         }
76         printf("Case %d: %d\n", ++kase, solve());
77     }
78     return 0;
79 }