ZOJ 1084 Channel Allocation（dfs）

Channel Allocation

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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. 

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Source: South Africa 2001

本题要求邻近的转发器使用不同的频道，同时转发器转发的信号是平面的区域，若把不同的频道看成着色问题，这就是著名的四色问题（呃。。。计算机证明的此猜想）

，即平面图中的着色最多四种。

实现：使用函数bool dfs(int id,int color)实现着色算法，id是起始着色的结点，color是限制的着色数量。当某一结点id使用颜色i(0<=i<color)时，判断前面已经着色的相邻结点是否有使用此颜色的，如果有，换颜色，否则，继续搜索下一个结点。当所有结点着色完毕，返回true；当颜色数color都用完了，还不能给所有结点着色，则返回false。

AC代码：

#include <stdio.h>
#include <memory.h>

bool g[26][26];
int used[26];
int n;

bool dfs(int id, int color)
{
	int i, j;
	bool flag;
	for (i = 0; i < color; i++)
	{
		flag = true;
		used[id] = i;
		for (j = 0; j < id; j++)//判断以前的相邻结点是否使用了该颜色 
		{
			if (g[id][j] && used[id] == used[j])
			{
				flag = false;
				break;
			}
		}
		if (flag && id == n-1)
			return true;
		if (flag && dfs(id+1,color))
		    return true;
	}
	//使用color个颜色没有实现全部着色 
	return false;
}

int main()
{
	int i, j;
	bool one;
	char adjacent[50];
	while (scanf("%d", &n) && n)
	{
		memset(g, 0, sizeof(g));
		memset(used, 0, sizeof(used));
		one = true;
		for ( i = 0; i < n; i++)  {
			scanf("%s", adjacent);
			for (j = 2; adjacent[j]; j++)
			{
				g[i][adjacent[j] - 'A'] = true;
				g[adjacent[j] - 'A'][i] = true;
				one = false;//存在相邻，一种颜色显然不可以 
			}
		}
		if (one)
			printf("1 channel needed.\n");
		else if (dfs(1, 2))
			printf("2 channels needed.\n");
		else if (dfs(1, 3))
			printf("3 channels needed.\n");
		else
			printf("4 channels needed.\n");
	}
	return 0;
}