poj1129 Channel Allocation

题意：

给一个图染色，相邻的点不能染相同的颜色，问最少需要多少种颜色。

思路：

搜索。

注意：在给一个新点染色的时候，用过的颜色中可以用的要试一下，没用过的颜色也要试一下，才不会出错。

数据：

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.

实现：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 30, INF = 0x3f3f3f3f;

int G[MAXN][MAXN], color[MAXN], n, ans;
string s;

void dfs(int now, int c)
{
    if (now == n)
    {
        ans = min(ans, c);
        return;
    }
    int * buf = new int[MAXN];
    for (int i = 0; i < MAXN; i++) buf[i] = 0;
    int maxn = 0;
    for (int i = 0; i < now; i++)
    {
        if (G[now][i] && color[i])
        {
            maxn = max(maxn, color[i]);
            buf[color[i]] = 1;
        }
    }
    for (int i = 1; i <= maxn + 1; i++)
    {
        if (!buf[i])
        {
            color[now] = i;
            dfs(now + 1, max(i, c));
        }
    }
    delete buf;
}

int main()
{
    while (cin >> n, n)
    {
        memset(G, 0, sizeof(G));
        memset(color, 0, sizeof(color));
        ans = INF;
        for (int i = 0; i < n; i++)
        {
            cin >> s;
            int l = s.length();
            for (int j = 2; j < l; j++)
            {
                int tmp = s[j] - 'A';
                G[i][tmp] = G[tmp][i] = 1;
            }
        }
        dfs(0, 0);
        cout << ans << " channel" << (ans > 1 ? "s" : "") << " needed." << endl;    
    }
    return 0;
}