快速切题 poj1129 Channel Allocation
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 12334 | Accepted: 6307 |
Description
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
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
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. 实际用时:21min
原因:....心理准备
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn=50;
int n;
vector <int > G[maxn];
int color[maxn];
int cnt;
bool dfs(int s,int c){
color[s]=c;
for(int i=0;i<G[s].size();i++){
int to=G[s][i];
if(color[to]==c){color[s]=-1;return false;}
if(color[to]==-1){
for(int i=0;i<=cnt;i++){
if(i==cnt)cnt++;
if(i!=c&&dfs(to,i))break;
}
}
}
return true;
}
char buff[50];
int main(){
while(scanf("%d",&n)==1&&n){
gets(buff);
for(int i=0;i<n;i++){
gets(buff);
G[i].clear();
for(int j=2;buff[j];j++){
G[i].push_back(buff[j]-'A');
}
}
memset(color,-1,sizeof(color));
cnt=1;
for(int i=0;i<n;i++)if(color[i]==-1)dfs(i,0);
if(cnt==1)printf("1 channel needed.\n");
else printf("%d channels needed.\n",cnt);
}
return 0;
}
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