POJ 1129 Channel Allocation（DFS）

Channel Allocation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13173		Accepted: 6737

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.

     题意：平面内最多26个点，给出一些点与点之间的矛盾关系。问最少使用多少颜

色才干给这些点染色，并保证矛盾点之间不同色。


#include<iostream>

#include<algorithm>

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<vector>

using namespace std;

int n;

int v[30];

char str[31];

int minn;

vector<int>q[30];

int up(int x,int k)

{

    for(int i=0;i<q[x].size();i++)

    {

        if(v[q[x][i]] == k)

        {

            return -1;

        }

    }

    return 1;

}

void DFS(int m,int num)

{

    if(m == n)

    {

        minn = min(minn,num);

        return ;

    }

    for(int i=0;i<num;i++)

    {

        v[m] = i;

        if(up(m,i) == 1)

        {

            DFS(m+1,num);

        }

    }

    v[m] = num;

    DFS(m+1,num+1);

    v[m] = -1;

}

int main()

{

    while(scanf("%d",&n)!=EOF)

    {

        if(n == 0)

        {

            break;

        }

        minn = 999999;

        for(int i=0;i<=30;i++)

        {

            q[i].clear();

        }

        for(int i=0;i<n;i++)

        {

            scanf("%s",str);

            for(int j=2;str[j]!='\0';j++)

            {

                q[(str[0]-'A')].push_back(str[j]-'A');

            }

        }

        for(int i=0;i<30;i++)

        {

            v[i] = -1;

        }

        DFS(0,0);

        if(minn == 1)

        {

            printf("1 channel needed.\n");

        }

        else

        {

            printf("%d channels needed.\n",minn);

        }

    }

    return 0;

}

巴特西

POJ 1129 Channel Allocation（DFS）

最新文章

热门文章