ZOJ3516 (图的遍历）

Tree of Three

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Now we have a tree and some queries to deal with. Every node in the tree has a value on it. For one node A, we want to know the largest three values in all the nodes of the subtree whose root is node A. Node 0 is root of the tree, except it, all other nodes have a parent node.

Input

There are several test cases. Each test case begins with a line contains one integer n(1 ≤ n ≤ 10000), which indicates the number of the node in the tree. The second line contains one integer v[0], the value on the root. Then for the following n - 1 lines(from the 3rd line to the (n + 1)th line), let i + 2 be the line number, then line i + 2 contains two integers parent and v[i], here parent is node i's parent node, v[i] is the value on node i. Here 0 ≤ v[i] ≤ 1000000. Then the next line contains an integer m(1 ≤ m ≤ 10000), which indicates the number of queries. Following m lines, each line contains one integer q, 0 ≤ q < n, it meas a query on node q.

Output

For each test case, output m lines, each line contains one or three integers. If the query asked for a node that has less than three nodes in the subtree, output a "-1"; otherwise, output the largest three values in the subtree, from larger to smaller.

Sample Input

5
1
0 10
0 5
2 7
2 8
5
0
1
2
3
4

Sample Output

10 8 7
-1
8 7 5
-1
-1

建图方法及遍历：链式齐向前星。
http://blog.csdn.net/acdreamers/article/details/16902023(链式向前星)

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<cstdlib>
 using namespace std;
 int num,k,pa;
 #define maxn 32010
 int val[maxn];
 int head[maxn];
 int cnt[maxn];
 struct Edge
 {
     int u;
     int v;
     int next;
 };
 Edge edge[maxn];
 void addedge(int u,int v)//链式前向星模板
 {
     edge[num].u=u;
     edge[num].v=v;
     edge[num].next=head[u];
     head[u]=num++;
 }
 void dfs(int p)
 {
   for(int i=head[p];i!=-;i=edge[i].next)
   {
       int v=edge[i].v;
       cnt[k++]=val[v];
       dfs(v);
   }
 }
 int main()
 {
     int n,root;
     while(~scanf("%d",&n))
     {
         memset(head,-,sizeof(head));
         num=;
         scanf("%d",&val[]);
         for(int i=;i<n;i++)
         {
             scanf("%d%d",&pa,&val[i]);
             addedge(pa,i);
         }
         int m,pos;
         scanf("%d",&m);
         for(int i=;i<m;i++)
         {
             k=;
             memset(cnt,,sizeof(cnt));
             scanf("%d",&pos);
             cnt[k++]=val[pos];
             dfs(pos);
             if(k<)
             printf("-1\n");
             else
             {
                 sort(cnt,cnt+k);
                 reverse(cnt,cnt+k);
                 printf("%d %d %d\n",cnt[],cnt[],cnt[]);
             }
         }
     }

     return ;
 }