首页 > 代码库 > poj1129 Channel Allocation 染色问题

poj1129 Channel Allocation 染色问题

2024-07-20 00:12:02   218人阅读

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. 问题模型：给出一幅连通图 每个字母代表一块区域 每行的冒号右边代表与该区域连通的区域    要求相邻的区域填充不同的颜色  问最少填几种颜色代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int map[30][30];         //是否连通
int color[30];           //该区域填的第几种颜色
int ans; 
int tt,t;                
bool ok(int a,int b)
{
    for(int i=0;i<t;i++)
    {
        if(map[i][a]==1&&b==color[i])          //与其连通的区域 如果有第b种颜色  则该区域不能填第b种颜色
            return false;
    }
    return true;
}
void dfs(int a,int colornum)
{
    if(tt==1)
        return;
    if(a>=t)
    {
        tt=1;
        return;
    }
    for(int i=1;i<=colornum;i++)
    {
        if(ok(a,i))
         {
             color[a]=i;
        dfs(a+1,colornum);            //填下一区域
            color[a]=0;
         }
    }
    if(tt==0)
    {
        ans++;
        dfs(a,colornum+1);             //未填充完 加一种颜色
    }
}
int main()
{
    char q[55];
    int m,n,i,j,k;
    while(cin>>t&&t)
    {
        tt=0;
        memset(color,0,sizeof(color));
        memset(map,0,sizeof(map));
        for(j=1;j<=t;j++)
        {
            scanf("%s",q);
            for(i=2;i<strlen(q);i++)
            {
                map[q[0]-'A'][q[i]-'A']=1;
            }
        }
        ans=1;
        dfs(0,1);
        if(ans==1)
            printf("1 channel needed.\n");//注意单复数
        else
            printf("%d channels needed.\n",ans);
    }
    return 0;
}



暴力搜索。。

poj1129 Channel Allocation 染色问题


联系
我们