POJ 1129-Channel Allocation（顶点着色）

- 日理万妓 2022-07-17 02:28 128阅读 0赞

Channel Allocation

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;14559</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;7406</td> 
  </tr> 
 </tbody> 
</table>

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][]

## 题目意思： ##

N个中继器，随后的N行中，A： 后面是与A相邻的中继器编号。

对于一个给定的中继器网络，所使用频道数应当尽可能地少，计算使用最少频道的数目。

## 解题思路： ##

顺序着色法求解。

无向图，染色，判断一共用了多少种颜色。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <vector>
    #include <queue>
    #include <map>
    #include <algorithm>
    using namespace std;
    #define INF 0xfffffff
    #define maxn 100010
    char s[26];//读入
    int edge[26][26];//邻接矩阵
    int n;//中继器数目
    int ans,c[26];//顶点着色数、各个顶点的颜色
    int b[26];//1表示该种颜色已经被使用
    void greedy()
    {
        int i,j;
        for(i=0; i<n; ++i)//给第i个顶点着色
        {
            memset(b,0,sizeof(b));
            for(j=0; j<n; ++j)//检查顶点i的每个邻接顶点
                if(edge[i][j]&&c[j]!=-1)//邻接顶点j已经着色且颜色为c[k]
                    b[c[j]]=1;
            for(j=0; j<=i; ++j)//j是顶点i的所有邻接顶点中未使用的且编号最小的颜色
                if(!b[j]) break;
            c[i]=j;//顶点i着第j种颜色
        }
        for(i=0; i<n; ++i)
            if(ans<c[i]) ans=c[i];
        ++ans;
    }
    int main()
    {
        int i,j;
        while(cin>>n,n)
        {
            memset(edge,0,sizeof(edge));
            for(i=0; i<n; ++i) c[i]=-1;
            ans=0;
            for(i=0; i<n; ++i)
            {
                cin>>s;
                int m=strlen(s)-2;
                for(j=0; j<m; ++j)//根据相邻关系建图
                {
                    edge[i][s[j+2]-'A']=1;
                    edge[s[j+2]-'A'][i]=1;
                }
            }
            greedy();
            if(ans!=1) cout<<ans<<" channels needed."<<endl;//注意ans=1要特判
            else cout<<ans<<" channel needed."<<endl;
        }
    }
    /**
    2
    A:
    B:
    4
    A:BC
    B:ACD
    C:ABD
    D:BC
    4
    A:BCD
    B:ACD
    C:ABD
    D:ABC
    0
    **/

[Southern African 2001]: http://poj.org/searchproblem?field=source&key=Southern+African+2001