poj 1129 Channel Allocation

Channel Allocation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9898		Accepted: 5059

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. 

Source

Southern African 2001

解题思路：

对于这题数据范围很小（节点最多26个），所以使用普通的暴力搜索法即可

对点i的染色操作：从最小的颜色开始搜索，当i的直接相邻（直接后继）结点已经染过该种颜色时，搜索下一种颜色。

就是说i的染色，当且仅当i的临近结点都没有染过该种颜色，且该种颜色要尽可能小。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct node
{
  int next[54];
  int pn;
};
int main( )
{
  int  n;
  int i,j;
  char s[54];
  while(scanf("%d",&n)!=EOF)
  {
     if(n==0) break;
     node point[n+n];
     int len;
     for( i=1;i<=n;i++)
      point[i].pn=0;
    
     for( i=1;i<=n;i++)
     {
    
       scanf("%s",s);len=strlen(s);
       for( j=2;j<len;j++)
       {
     
      int k=s[j]%('A'-1);////把结点字母转换为相应的数字，如A->1  C->3   
      point[i].next[++point[i].pn]=k;
       }
      
     }
     int color[54]={0};////color[i]为第i个结点当前染的颜色，0为无色（无染色）   
     color[1]=1;//结点A初始化染第1种色  
     int maxcolor=1; //当前已使用不同颜色的种数   
     for( i=1;i<=n;i++)//枚举每个顶点   
     {
      color[i]=n+1;//先假设结点i染最大的颜色
      int vis[54]={0};//标记第i种颜色是否在当前结点的相邻结点染过  
      for( j=1;j<=point[i].pn;j++)//枚举顶点i的所有后继
      {
      int k=point[i].next[j];
       if(color[k])//顶点i的第j个直接后继已染色   
       {
         vis[color[k]]=1;//标记该种颜色
    }
    }
  
    for( j=1;;j++)//从最小的颜色开始，枚举每种颜色，最终确定结点i的染色   
    {
     if(!vis[j]&&color[i]>j)
     {
      color[i]=j;
      break;
     
   }
       }
       if(maxcolor<color[i])  maxcolor=color[i];
    }
     printf("%d ",maxcolor);
     if(maxcolor==1) puts("channel needed.");
     else puts("channels needed.");
    
  }
  return 0;
}

链接:http://poj.org/problem?id=1129