POJ 1129：Channel Allocation 四色定理+暴力搜索

Channel Allocation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13357		Accepted: 6836

Description

When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a strong signal. However, the channels used by each repeater must be carefully chosen so that nearby repeaters do not interfere with one another. This condition is satisfied if adjacent repeaters use different channels. 

Since the radio frequency spectrum is a precious resource, the number of channels required by a given network of repeaters should be minimised. You have to write a program that reads in a description of a repeater network and determines the minimum number of channels required.

Input

The input consists of a number of maps of repeater networks. Each map begins with a line containing the number of repeaters. This is between 1 and 26, and the repeaters are referred to by consecutive upper-case letters of the alphabet starting with A. For example, ten repeaters would have the names A,B,C,...,I and J. A network with zero repeaters indicates the end of input. 

Following the number of repeaters is a list of adjacency relationships. Each line has the form: 

A:BCDH 

which indicates that the repeaters B, C, D and H are adjacent to the repeater A. The first line describes those adjacent to repeater A, the second those adjacent to B, and so on for all of the repeaters. If a repeater is not adjacent to any other, its line has the form 

A: 

The repeaters are listed in alphabetical order. 

Note that the adjacency is a symmetric relationship; if A is adjacent to B, then B is necessarily adjacent to A. Also, since the repeaters lie in a plane, the graph formed by connecting adjacent repeaters does not have any line segments that cross. 

Output

For each map (except the final one with no repeaters), print a line containing the minumum number of channels needed so that no adjacent channels interfere. The sample output shows the format of this line. Take care that channels is in the singular form when only one channel is required.

Sample Input

2
A:
B:
4
A:BC
B:ACD
C:ABD
D:BC
4
A:BCD
B:ACD
C:ABD
D:ABC
0

Sample Output

1 channel needed.
3 channels needed.
4 channels needed. 

题意是给了一堆的信号辐射塔，然后给了各个塔之间的关系，A:BCD代表A与B塔，A与C塔，A与D塔相邻，这样的话A就不能与B、C、D塔中的任何一个使用相同的辐射频率。问在给定关系的情况下，最少使用多少种辐射频率。

和小时候接触的世界地图的染色方案一样，相邻国家不能使用同一种颜色。这样最多就4种颜色。

这道题折磨了我颇久，用dfs一直不对，结果看了discuss发现顺序染色是不对的。

看给定数据：

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

3 channels needed.
即：A(1)B(2)C(3)D(1)E(2)F(3)

贪心做法是不对的，还是暴力搜吧，反正一共就26个点，四种颜色换着法地涂，看最后最少能用多少种颜色。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int n, maxn;
string test;
int grid[30][30];

void dfs(int step,int color[], int cap)
{
	if (step == n )
	{
		maxn = min(maxn, cap);
		return;
	}
	int i, j;
	for (i = 1; i <= 4; i++)//根据四色定理可知，一共四种颜色。所以暴力搜索，把所有符合条件的颜色都染着试一下
	{
		for (j = 0; j < 26; j++)
		{
			if (grid[step][j] == 1 && color[j] == i)
				break;
		}
		if (j == 26)//如果这个颜色符合条件
		{
			color[step] = i;
			dfs(step + 1, color, max(cap, i));
			color[step] = 0;
		}
	}
}

int main()
{
	int i, j, k, len;
	int color[30];
	while (cin >> n)
	{
		if (n == 0)
			break;
		memset(grid, 0, sizeof(grid));
		memset(color, 0, sizeof(color));
		maxn = 10;

		for (i = 0; i < n; i++)
		{
			cin >> test;
			len = test.length();

			if (len == 2)
				continue;
			k = test[0] - 'A';
			for (j = 2; j < len; j++)
			{
				grid[k][test[j] - 'A'] = 1;
				grid[test[j] - 'A'][k] = 1;
			}
		}
		dfs(0,color,0);

		if (maxn == 1)
		{
			cout <<"1 channel needed." << endl;
		}
		else
		{
			cout << maxn <<" channels needed. "<< endl;
		}
	}
	return 0;
}

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