POJ 1129 Channel Allocation(DFS)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 13173 | Accepted: 6737 |
Description
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
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
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;
}
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