hdu 4337 King Arthur's Knights
King Arthur's Knights
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 248 Accepted Submission(s): 109
Special Judge
- from Fate / Stay Night
You must have known the legend of King Arthur and his knights of the round table. The round table has no head, implying that everyone has equal status. Some knights are close friends with each other, so they prefer to sit next to each other.
Given the relationship of these knights, the King Arthur request you to find an arrangement such that, for every knight, his two adjacent knights are both his close friends. And you should note that because the knights are very united, everyone has at least half of the group as his close friends. More specifically speaking, if there are N knights in total, every knight has at least (N + 1) / 2 other knights as his close friends.
AC代码:
#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;
int map[200][200],rec[200],vis[200];
int flag ,n,m;
vector<int>v[200];
void dfs( int x,int k)
{
if(flag) return;
rec[k]=x;
int i;
if(k==n&&map[x][1])
{
flag=1;
for( i=1;i<=n;i++)
{
printf(i==1?"%d":" %d",rec[i]);
}
puts("");
return ;
}
int len=v[x].size( );
for( i=0;i<len;i++)
{
if(vis[v[x][i]]==0)
{
vis[v[x][i]]=1;
dfs(v[x][i],k+1);
vis[v[x][i]]=0;
}
}
}
int main( )
{
int x,y;
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(vis,0,sizeof(vis));
memset(rec,0,sizeof(rec));
memset(map,0,sizeof(map));
for( i=0;i<m;i++)
{
scanf("%d%d",&x,&y);
map[x][y]=map[y][x]=1;
v[x].push_back(y);
v[y].push_back(x);
}
flag=0;
vis[1]=1;
dfs(1,1);
for( i=1;i<=n;i++)
v[i].clear( );
}
return 0;
}