poj1129Channel Allocation 乱做，dfs或+四色定理

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

1. 题意提炼：无向图任意相邻的顶点之间涂不同的颜色，要求所使用颜色种类最少
2. 分析：每图一个点只能确定其相邻的点不能图哪种颜色，不能确定不相邻的点是否可以涂此种颜色，所以记录每个点不能涂哪几种颜色
3. 每次从编号最小的颜色开始尝试是否可以涂，确定后更新其他相邻顶点不能涂此种颜色
4. 代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <queue>#include <vector>#include <stack>using namespace std;int n;int color[30];//每个点的颜色vector<int> no[30];//每个点不能涂的颜色vector<int> map[30];//图，后来发现没什么用，因为处理合并到读入里了int main(){    char ch;    int p,q;    int col;    while(scanf("%d",&n),n!=0){        //clear        memset(color,0,sizeof(color));        for(int i=0;i<n;i++){            no[i].clear();            map[i].clear();        }        //read        getchar();        for(int i=0;i<n;i++){            ch=getchar();            p=ch-'A';//index            //            col=1;            bool flag=true;//continu find color        //  cout<<"find color "<<p<<endl;            while(flag){                flag=false;                for(int j=0;j<no[p].size();j++){//color                    if(col==no[p][j]){//ijyaoxiedui a                         //cout<<"no "<<col<<" ";                        col++;                        flag=true;                        break;                    }                }            }            color[p]=col;            //            getchar();            while(ch=getchar(),ch!='\n'){                q=ch-'A';                map[p].push_back(q);                no[q].push_back(col);            }        }        int maxx=0;        for(int i=0;i<n;i++){        //  cout<<color[i]<<" ";            if(color[i]>maxx)maxx=color[i];        }        maxx==1?printf("%d channel needed.\n",maxx):printf("%d channels needed.\n",maxx);    }    return 0;}

其他DFS的方法

题目翻译+四色定理