题目传送门
1 /*
2 模拟水题，排序后找出重复的ip就可以了
3 */
4 #include <cstdio>
5 #include <iostream>
6 #include <algorithm>
7 #include <cstring>
8 #include <string>
9 #include <cmath>
10 using namespace std;
11
12 const int MAXN = 1e3 + 10;
13 const int INF = 0x3f3f3f3f;
14 struct Ip
15 {
16 int a, b, c, d;
17 }ip[MAXN];
18 struct H
19 {
20 int a, b, c, d;
21 }h[MAXN];
22 int ans[55];
23
24 bool cmp(H x, H y)
25 {
26 if (x.a == y.a)
27 {
28 if (x.b == y.b)
29 {
30 if (x.c == y.c) return x.d < y.d;
31 else return x.c < y.c;
32 }
33 else return x.b < y.b;
34 }
35 else return x.a < y.a;
36 }
37
38 int main(void) //2015百度之星资格赛 1003 IP聚合
39 {
40 int t, cas = 0;
41 scanf ("%d", &t);
42 while (t--)
43 {
44 int n, m; scanf ("%d%d", &n, &m);
45 for (int i=1; i<=n; ++i)
46 scanf ("%d.%d.%d.%d", &ip[i].a, &ip[i].b, &ip[i].c, &ip[i].d);
47
48 memset (ans, 0, sizeof (ans));
49 for (int i=1; i<=m; ++i)
50 {
51 int u, v, w, x;
52 scanf ("%d.%d.%d.%d", &u, &v, &w, &x);
53 for (int j=1; j<=n; ++j)
54 {
55 h[j].a = (ip[j].a & u); h[j].b = (ip[j].b & v);
56 h[j].c = (ip[j].c & w); h[j].d = (ip[j].d & x);
57 }
58
59 sort (h+1, h+1+n, cmp);
60
61 int cnt = 0;
62 for (int k=2; k<=n; ++k)
63 {
64 if (h[k].a == h[k-1].a &&
65 h[k].b == h[k-1].b &&
66 h[k].c == h[k-1].c &&
67 h[k].d == h[k-1].d) cnt++;
68 }
69
70 ans[i] = (n - cnt);
71 }
72
73 printf ("Case #%d:
", ++cas);
74 for (int i=1; i<=m; ++i) printf ("%d
", ans[i]);
75 }
76
77 return 0;
78 }
79
80 /*
81 2
82 5 2
83 192.168.1.0
84 192.168.1.101
85 192.168.2.5
86 192.168.2.7
87 202.14.27.235
88 255.255.255.0
89 255.255.0.0
90 4 2
91 127.127.0.1
92 10.134.52.0
93 127.0.10.1
94 10.134.0.2
95 235.235.0.0
96 1.57.16.0
97 */