POJ1129

为每个节点选一个频道，相邻的节点的频道不能相同。

采用深搜，为每个节点选一个编号，从1开始选，因为是从1开始选的，所以选到最后能匹配的一定是最小的频道数。

选一个节点的时候判断一下就OK了。

搜索的题目基本上在POJ上AC完了，UVA上的也A完了。现在对搜索的题目感觉挺有自信的。。呵呵

AC代码：

#include <iostream>
#include <cstring>
using namespace std;
const int MAX_NODES = 28;
int nodes;
int repeater_number[MAX_NODES];
int maps[MAX_NODES][MAX_NODES];
int channels_number;
void dfs(int cur) {
	if (cur == nodes)
		return ;
	else {
		int flag = 1;
		for (int i = 1; flag; i++) {
			int ok = 1;
			for (int j = 0; j < nodes; j++) {
				if (maps[cur][j] && repeater_number[j] == i){
					ok = 0;
					break;
				}
			}
			if (ok){
				repeater_number[cur] = i;
				dfs(cur+1);
				flag = 0;
			}
		}
	}
}
int main()
{
	while (cin >> nodes) {
		if (!nodes) 
			break;
		cin.get();
		memset(repeater_number,0,sizeof(repeater_number));
		memset(maps,0,sizeof(maps));
		for (int i = 0; i < nodes; i++) {
			char str[MAX_NODES];
			cin.getline(str,MAX_NODES);
			int source_node, target_node;
			source_node = str[0] - 'A';
			for (int j = 2; str[j]; j++) {
				target_node = str[j] - 'A';
				maps[source_node][target_node] = 1;
				maps[target_node][source_node] = 1;
			}
		}
		channels_number = 0;
		dfs(0);
		for (int i = 0; i < nodes; i++) {
			if(repeater_number[i] > channels_number)
				channels_number = repeater_number[i];
		}
		if (channels_number > 1) {				
			cout << channels_number << " channels needed." << endl;
		}
		else
			cout << channels_number << " channel needed." << endl;
	}
	return 0;
}