POJ 1129 Channel Allocation
Channel Allocation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 6071 | Accepted: 3108 |
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.
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.
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
1 #include <cstdlib> 2 #include <iostream> 3 #include <string.h> 4 #include <stdio.h> 5 6 using namespace std; 7 struct Node{ 8 int data[27]; 9 int rear; 10 }q[27]; 11 12 int n; 13 14 int data[27]; 15 int check(int x,int i) 16 {int j; 17 for(j=0;j<q[x].rear;j++) 18 if(data[q[x].data[j]]==i) 19 return 0; 20 21 return 1; 22 } 23 int mount; 24 int visited[27]; 25 int dfs(int x) 26 {int i,j,k; 27 int sum=0; 28 k=0; 29 for(j=1;j<=n;j++) 30 if(visited[j]) 31 k++; 32 if(x>n) 33 if(k<mount) 34 { mount=k;return 1;} 35 if(k>=mount) 36 return 0; 37 38 39 if(x>n) 40 {for(i=1;i<=n;i++) 41 if(visited[i]) 42 sum++; 43 //cout<<sum<<endl; 44 if(sum<mount) 45 mount=sum; 46 return 1; 47 } 48 for(i=1;i<=n;i++) 49 { 50 51 if(!check(x,i)) 52 continue; 53 54 55 data[x]=i; 56 visited[i]++; 57 dfs(x+1); 58 data[x]=0; 59 visited[i]--; 60 } 61 return 0; 62 63 } 64 65 66 int main(int argc, char *argv[]) 67 { 68 int i,j,k; 69 while(cin>>n,n) 70 {char a,b,c; 71 72 memset(data,0,sizeof(data)); 73 memset(visited,0,sizeof(visited)); 74 75 mount=45; 76 77 for(i=1;i<=n;i++) 78 {cin>>a; 79 cin>>c; 80 q[a-'A'+1].rear=0; 81 while(1) 82 { 83 b=getchar(); 84 if(b==10) 85 break; 86 q[a-'A'+1].data[q[a-'A'+1].rear++]=(int)(b-'A'+1); 87 88 } 89 } 90 dfs(1); 91 cout<<mount; 92 if(mount==1) 93 cout<<" channel needed."<<endl; 94 else 95 cout<<" channels needed."<<endl; 96 97 98 99 } 100 101 102 103 104 105 106 107 108 109 110 111 112 113 system("PAUSE"); 114 return EXIT_SUCCESS; 115 }
