poj 1129 Channel Allocation （DFS 图的染色问题）

Channel Allocation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13152 Accepted: 6721

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

2A:B:4A:BCB:ACDC:ABDD:BC4A:BCDB:ACDC:ABDD:ABC0

Sample Output

1 channel needed.3 channels needed.4 channels needed. 

Source

Southern African 2001

题目链接：http://poj.org/problem?id=1129

题目大意：给出一些不同的中继器（以对应的不同的大写字母表示），以及与它们邻接的中继器。相邻的中继器只能使用不同的频道，求最少使用的频率数。即典型的图的染色问题。

解题思路：由四色定理可知，无论多复杂的图，最多可由四种颜色完成图的染色问题。用邻接矩阵存储点与点之间的相邻关系。设置1,2,3,4为四个颜色值。所以枚举1-4为解，当一个解满足条件时即为答案。DFS递归过程中枚举每个点的颜色值并判断是否满足条件。和数独思想类似。

代码如下：

#include <cstdio>#include <cstring>int n;int a[30][30],c[30];char s[30];bool p;int col;bool ok(int x){for(int i=0;i<n;i++){if(a[x][i] && c[x]==c[i])return false;}return true;}void dfs(int x){if(p)return;if(x>=n){p=true;return ;}for(int j=1;j<=col;j++){ c[x]=j;if(ok(x)){dfs(x+1);if(p)return ;}c[x]=0;}}int main(){while(scanf("%d",&n)!=EOF && n){p=false;memset(a,0,sizeof(a));memset(c,0,sizeof(c));for(int i=0;i<n;i++){scanf("%s",s);int m=strlen(s);int x=s[0]-'A';for(int j=2;j<m;j++){int y=s[j]-'A';a[x][y]=1;a[y][x]=1;}}for(col=1;col<=4;col++){dfs(0);if(p)break;}if(col==1)printf("1 channel needed.\n");elseprintf("%d channels needed.\n",col);}return 0;}