Codeforces 598D (ccpc-wannafly camp day1) Igor In the Museum

http://codeforces.com/problemset/problem/598/D

分析：BFS，同一连通区域的周长一样，但查询过多会导致TLE，所以要将连通区域的答案储存，下次查询到该连通区域就可以直接得出结果

 1 #include<iostream>
 2 #include<sstream>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<string>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<functional>
 9 #include<iomanip>
10 #include<numeric>
11 #include<cmath>
12 #include<queue>
13 #include<vector>
14 #include<set>
15 #include<cctype>
16 const double PI = acos(-1.0);
17 const int INF = 0x3f3f3f3f;
18 const int NINF = -INF - 1;
19 typedef long long ll;
20 #define MOD 1000007
21 using namespace std;
22 typedef pair<int, int> P;
23 char map[1005][1005];
24 int n, m, ans, cnt;
25 int x, y;
26 int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
27 int vis[1005][1005];
28 int num[1000025];
29 void bfs()
30 {
31     int i, j;
32     queue<P> q;
33     q.push(P(x, y));
34     cnt++;
35     vis[x][y] = cnt;
36     ans = 0;
37     while (q.size())
38     {
39         P temp = q.front();
40         q.pop();
41         ans += 4;
42         for (i=0; i < 4; ++i)
43         {
44             int nx = temp.first + dx[i];
45             int ny = temp.second + dy[i];
46             if (nx >= 0 && nx < n && ny >= 0 && ny < m)
47             {
48                 if(map[nx][ny] == '.')
49                 {
50                     ans--;
51                     if(!vis[nx][ny])
52                     {
53                         q.push(P(nx, ny));
54                         vis[nx][ny] = cnt;
55                     }
56                 }
57             }
58             else
59                 ans--;
60         }
61     }
62     num[cnt] = ans;
63     cout << ans << endl;
64 }
65
66 int main()
67 {
68     int i, j, k;
69     cin >> n >> m >> k;
70     for (i = 0; i < n; ++i)
71         cin >> map[i];
72     memset(num, -1, sizeof(num));
73     memset(vis, 0, sizeof(vis));
74     cnt = 0;
75     while (k--)
76     {
77         cin >> x >> y;
78         x--, y--;
79         if (vis[x][y] == 0)
80             bfs();
81         else
82             cout << num[vis[x][y]] << endl;
83     }
84     return 0;
85 }