UVa 10054 项链

题意：

思路：

Code：

#include<stdio.h>
#include<string.h>

bool solve();
void dfs(int n);

int graph[51][51];
int du[51];
int path[1010*2];
int len;

int main()
{
  //freopen("10054.in","r",stdin);
  //freopen("10054.out","w",stdout);
  int t;
  scanf("%d",&t);
  int num=0;
  while(t-->0)
  {
    num++;
    memset(graph,0,sizeof(graph));
    memset(du,0,sizeof(du));
    len=0;
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;++i)
    {
      int a,b;
      scanf("%d%d",&a,&b);
      du[a]++;
      du[b]++;
      graph[a][b]++;//增加一条无向边
      //graph[b][a]=graph[a][b];
      graph[b][a]++;
    }
    if(num!=1) printf("\n");
    printf("Case #%d\n",num);
    int len=0;
    if(solve()==false) printf("some beads may be lost\n");
  }
  return 0;
}

bool solve()
{
  //判定顶点度数
  for(int i=1;i<=50;++i)
  {
    if(du[i]%2==0) continue;
    else return false;
  }//printf("ok\n");
  //判定连通性
  int cnt=0;//连通块个数
  for(int i=1;i<=50;i++)
  {
    if(du[i])//存在该颜色
    {
      cnt++; dfs(i);
    }
  }
  if(cnt>1) return false;
  //print
  //printf("len:%d\n",len);
  for(int i=0;i<len-1;i=i+2)
  {
    printf("%d %d\n",path[i],path[i+1]);
  }
  //printf("%d %d\n",path[len-1],path[0]);
  return true;
}

void dfs(int n)
{
  for(int i=1;i<=50;i++)
  {
    if(graph[n][i])
    {
      graph[n][i]--;
      //graph[i][n]=graph[n][i];
      graph[i][n]--;
      du[n]--;
      du[i]--;
      //path[len++]=n;
      dfs(i);
      path[len++]=i;
      path[len++]=n;
    }
  }
}