poj1129 Channel Allocation

題意：

給一個圖染色，相鄰的點不能染相同的顏色，問最少須要多少種顏色。

思路：

搜索。

注意：在給一個新點染色的時候，用過的顏色中能夠用的要試一下，沒用過的顏色也要試一下，纔不會出錯。

數據：

10
A:BCDEFG
B:ACGH
C:ABDH
D:ACEHJ
E:ADFIJ
F:AEGI
G:ABFHI
H:BCDGIJ
I:EFGHJ       
J:DEHI

4 channels needed.

6
A:BEF
B:AC
C:BD
D:CEF
E:ADF
F:ADE

3 channels needed.

實現：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 
 8 const int MAXN = 30, INF = 0x3f3f3f3f;
 9 
10 int G[MAXN][MAXN], color[MAXN], n, ans;
11 string s;
12 
13 void dfs(int now, int c)
14 {
15     if (now == n)
16     {
17         ans = min(ans, c);
18         return;
19     }
20     int * buf = new int[MAXN];
21     for (int i = 0; i < MAXN; i++) buf[i] = 0;
22     int maxn = 0;
23     for (int i = 0; i < now; i++)
24     {
25         if (G[now][i] && color[i])
26         {
27             maxn = max(maxn, color[i]);
28             buf[color[i]] = 1;
29         }
30     }
31     for (int i = 1; i <= maxn + 1; i++)
32     {
33         if (!buf[i])
34         {
35             color[now] = i;
36             dfs(now + 1, max(i, c));
37         }
38     }
39     delete buf;
40 }
41 
42 int main()
43 {
44     while (cin >> n, n)
45     {
46         memset(G, 0, sizeof(G));
47         memset(color, 0, sizeof(color));
48         ans = INF;
49         for (int i = 0; i < n; i++)
50         {
51             cin >> s;
52             int l = s.length();
53             for (int j = 2; j < l; j++)
54             {
55                 int tmp = s[j] - 'A';
56                 G[i][tmp] = G[tmp][i] = 1;
57             }
58         }
59         dfs(0, 0);
60         cout << ans << " channel" << (ans > 1 ? "s" : "") << " needed." << endl;    
61     }
62     return 0;
63 }