POJ-1129 Channel Allocation （DFS）

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. 

题目大意：一张图中，相邻两点不能涂同一种颜色，要把整张图都涂上，最少需要多少种颜色。
题目解析：数据量不大，DFS即可。根据四色原理，最多只有四种颜色。先将第一个点涂上一种颜色（这是必然的），然后一个点一个点涂下去，当涂到最后一个点时，合计当前方案的颜色种数，然后更新最优解。

代码如下：

 # include<iostream>
 # include<cstdio>
 # include<set>
 # include<string>
 # include<cstring>
 # include<algorithm>
 using namespace std;
 int n,ans,col[],mp[][];
 bool ok(int p,int c)
 {
     for(int i=;i<n;++i)
         if(mp[p][i]&&col[i]==c)
             return false;
     return true;
 }
 void dfs(int p)
 {
     if(p==n-){
         for(int k=;k<=;++k){
             if(ok(p,k)){
                 col[p]=k;
                 set<int>s;
                 for(int i=;i<n;++i){
                     s.insert(col[i]);
                 }
                 if(ans>s.size())
                     ans=s.size();
                 col[p]=;
             }
         }
         return ;
     }
     for(int i=;i<=;++i){
         if(ok(p,i)){
             col[p]=i;
             dfs(p+);
             col[p]=;
         }
     }
 }
 int main()
 {
     while(scanf("%d",&n)&&n)
     {
         string p;
         memset(mp,,sizeof(mp));
         for(int i=;i<n;++i){
             cin>>p;
             for(int j=;j<p.size();++j)
                 mp[i][p[j]-'A']=mp[p[j]-'A'][i]=;
         }
         memset(col,,sizeof(col));
         ans=;
         col[]=;
         dfs();
         if(ans==){
             printf("%d channel needed.\n",ans);
         }else
             printf("%d channels needed.\n",ans);
     }
     return ;
 }