poj 1129 也算是遍历的吧 两种方法

     我当时是不懂题目是什么意思的，看了别人的才知道原来是所谓的四色问题。我也不知道是什么东东，查了才懂。要学的还有很多呢！太弱了。关键把结构存储进去后，每次遍历它的邻接节点。并把用的颜色标记下来。然后再每次都要遍历每种颜色，如果这种颜色不是邻接的，那么久可以把它给用上。用col[]数组来表示总共用了多少种颜色。好了，代码里也标记得很清楚了。这题还有其它方法，要好好研究研究。  感谢http://blog.csdn.net/lyy289065406。推荐大家看他的博客。

                                                                                   Channel Allocation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9268		Accepted: 4711

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

#include <iostream>
#include <fstream>

using namespace std;

bool vis[27];
int col[27];  //节点对应使用的颜色 col[i]=j.表示节点i使用了颜色j。
typedef class{
    public:
      int next[27];
      int pn;  
}point;

int n;

void getmap()
{
 


}
int main()
{
    freopen("acm.txt","r",stdin);
    while(scanf("%d",&n) && n)
    {
        int i,j,maxcolor=1;
        memset(col,0,sizeof(col));
        point *node=new point[n+1];
        
        for(i=1;i<=n;i++)  
        {  
           getchar();  //吸收第一个字母   
           getchar();  //吸收冒号   
          if(node[i].pn<0)   //初始化指针   
            node[i].pn=0;  
          char ch;  
          while((ch=getchar())!='\n')  
          {   
            j=ch%('A'-1);   //把结点字母转换为相应的数字，如A->1  C->3   
            node[i].next[ ++node[i].pn ]=j;  
          }  
        }

        for(i=1; i<=n; i++)
        {
           col[i]=n+1;
           memset(vis,false,sizeof(vis));
           for(j=1; j<=node[i].pn; j++)  //判断某种颜色是否被邻接节点使用
           {
               if(col[node[i].next[j]])  
                  vis[ col[node[i].next[j]] ]=true; //把这种颜色标记一下
           }
           for(j=1; j<=n; j++)  //遍历每种颜色
           {
               if(!vis[j] && col[i]>j)  //某种颜色没有被邻接节点使用
               {
                   col[i]=j;
                   break;
               }
           }
           if(maxcolor<col[i])
              maxcolor=col[i];
        }
        if(maxcolor==1)
        {
            printf("%d channel needed.\n",maxcolor);
        }
        else
        {
           printf("%d channels needed.\n",maxcolor);
        }
        delete node;
    }
    return 0;
}

 

DFS：

#include <iostream>
#include <fstream>
#include <memory.h>

using namespace std;

bool map[27][27];
int color[27],color_count;
int n,mincolor,flag;
char s[27];

bool isOK(int dept,int col)
{   //判断某种颜色是否合适
    for(int i=1; i<=n ;i++)
    {  
        if(map[dept][i] && color[i]==col)
        {
            return false;
        }
    }
    return true;
}

void dfs(int depth)
{
    if(flag) return;
    if(depth==n+1)
    {  
        mincolor=color_count;
        flag=1;
    }
    for(int j=1; j<=color_count; j++) //判断用过的颜色里是否有可用的
    {
        if(isOK(depth,j))
        {
          color[depth]=j;
          dfs(depth+1);
          color[depth]=0;
        }
    }

    //选用一种新的颜色
    color_count++;
    color[depth]=color_count;
    dfs(depth+1);
    color[depth]=0;
    color_count--;
}

int main(){
  
    freopen("acm.txt","r",stdin);
   while(scanf("%d",&n)!=EOF && n)
   {
       memset(map,0,sizeof(map));
       //get the map
      for(int i=1; i<=n; i++)
      {
          getchar();
          scanf("%s",s);
          for(int j=2; s[j]!='\0'; j++)
              map[i][s[j]-'A'+1]=true;
      }
      mincolor=999;  flag=0; color_count=1;
      memset(color,0,sizeof(color));
      dfs(1);
      if(mincolor==1)
          printf("%d channel needed.\n",mincolor);
      else
          printf("%d channels needed.\n",mincolor);
   }
   return 0;
}