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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> Channel Allocation (poj 1129 dfs)

Channel Allocation (poj 1129 dfs)

編輯:C++入門知識

Channel Allocation (poj 1129 dfs)


Language: Channel Allocation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12367 Accepted: 6325

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

題意:給你一個n,代表電台的數量。電台的編號是從A到Z。然後給你他們之間的鄰接關系,讓你求出最小需要的頻率數。要求任意兩個相鄰的電台之間不允許用同一頻率。

思路:數據不大,最多26,dfs暴力,用鄰接表存圖,color[x]=i表示x號電台使用i頻率。

代碼:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 30
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

struct Edge
{
    int u,v;
    int next;
}edge[maxn*maxn];

int N,edgenum;
int head[maxn];
int color[maxn];

void addedge(int u,int v)
{
    edge[edgenum].v=v;
    edge[edgenum].next=head[u];
    head[u]=edgenum++;
}

bool ISok(int x)
{
    for (int i=head[x];i!=-1;i=edge[i].next)
    {
        if (color[x]==color[edge[i].v])
            return false;
    }
    return true;
}

bool dfs(int point_num,int color_num)
{
    if (point_num>N)
        return true;
    for (int i=1;i<=color_num;i++)
    {
        color[point_num]=i;
        if (ISok(point_num))
        {
            if (dfs(point_num+1,color_num))
                return true;
        }
        color[point_num]=0;
    }
    return false;
}

int main()
{
    while (scanf("%d",&N)&&N)
    {
        getchar();
        memset(head,-1,sizeof(head));
        memset(color,0,sizeof(color));
        edgenum=0;
        for (int i=1;i<=N;i++)
        {
            getchar();
            getchar();
            char ch;
            while (ch=getchar())
            {
                if (ch=='\n')
                    break;
                addedge(i,ch-'A'+1);
            }
        }
        for (int i=1;i<=N;i++)//從1~N枚舉顏色種類數
        {
            if (dfs(1,i))
            {
                if (i==1)//注意一個頻率是用channel,多個時用channels
                    printf("1 channel needed.\n");
                else
                    printf("%d channels needed.\n",i);
                break;
            }
        }
    }
    return 0;
}
/*
2
A:
B:
4
A:BC
B:ACD
C:ABD
D:BC
4
A:BCD
B:ACD
C:ABD
D:ABC
0
*/


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