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!=-1;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,-1,sizeof(head));
        num=0;
        scanf("%d",&val[0]);
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&pa,&val[i]);
            addedge(pa,i);
        }
        int m,pos;
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            k=0;
            memset(cnt,0,sizeof(cnt));
            scanf("%d",&pos);
            cnt[k++]=val[pos];
            dfs(pos);
            if(k<3)
            printf("-1
");
            else
            {
                sort(cnt,cnt+k);
                reverse(cnt,cnt+k);
                printf("%d %d %d
",cnt[0],cnt[1],cnt[2]);
            }
        }
    }

    return 0;
}