CodeForces 29D - Ant on the Tree 暴力LCA

         题意:

                    给一颗树...问能否从根节点出发回到根结点..并且每条边exactly经过两次..并且遍历叶子的顺序为所给的顺序..如果可以输出遍历路径..否则输出-1

         题解:

                    数据范围很小(N<=300)....所以就用超级暴力O(n^2)来水了...我首先将每个叶子结点的路径找出..然后暴力做LCA...相当于模拟遍历过程..并用一个计数器记录每条边访问的次数..LCA以前写过..正规的方法一定要捡起来..

Program:

#include
#include
#include
#include
#include
#include
#include
#define oo 1000000007
#define ll long long 
#define MAXN 305
using namespace std; 
struct node
{
       int way[MAXN],num;
}L[MAXN];
vector T[MAXN];
int way[MAXN],times[MAXN][MAXN],ans[2*MAXN],num,leaf;
bool anc[MAXN];
void dfs(int x,int father,int n)
{
       int i,m=T[x].size();
       way[n]=x;
       if (x!=1 && m==1)
       {
              ++leaf;
              L[x].num=n;
              for (i=1;i<=n;i++) L[x].way[i]=way[i];
              return;
       }
       for (i=0;i=1;i--) 
              {
                      if (times[L[pre].way[i+1]][L[pre].way[i]]>1) return false;
                      times[L[pre].way[i+1]][L[pre].way[i]]++;
                      ans[++num]=L[pre].way[i];
                      if (anc[L[pre].way[i]]) break;
              }
              if (pre==1) t=1;
                 else t=L[pre].way[i];
              for (i=1;i<=L[x].num;i++) 
                 if (L[pre].way[i]==t) break; 
              i++;
              for (;i<=L[x].num;i++)
              { 
                      if (times[L[x].way[i-1]][L[x].way[i]]>1) return false;
                      times[L[x].way[i-1]][L[x].way[i]]++;
                      ans[++num]=L[x].way[i];
              }
              pre=x;
       } 
       for (i=L[x].num-1;i>=1;i--)
       { 
              if (times[L[x].way[i+1]][L[x].way[i]]) return false;
              times[L[x].way[i+1]][L[x].way[i]]++;
              ans[++num]=L[x].way[i]; 
       }
       return true;
}
int main()
{  
       int n,i;
       freopen("input.txt","r",stdin);
       freopen("output.txt","w",stdout);  
       scanf("%d",&n);
       for (i=1;i<=n;i++) T[i].clear();
       for (i=1;i