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)
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstdlib> 6 using namespace std; 7 int num,k,pa; 8 #define maxn 32010 9 int val[maxn]; 10 int head[maxn]; 11 int cnt[maxn]; 12 struct Edge 13 { 14 int u; 15 int v; 16 int next; 17 }; 18 Edge edge[maxn]; 19 void addedge(int u,int v)//链式前向星模板 20 { 21 edge[num].u=u; 22 edge[num].v=v; 23 edge[num].next=head[u]; 24 head[u]=num++; 25 } 26 void dfs(int p) 27 { 28 for(int i=head[p];i!=-1;i=edge[i].next) 29 { 30 int v=edge[i].v; 31 cnt[k++]=val[v]; 32 dfs(v); 33 } 34 } 35 int main() 36 { 37 int n,root; 38 while(~scanf("%d",&n)) 39 { 40 memset(head,-1,sizeof(head)); 41 num=0; 42 scanf("%d",&val[0]); 43 for(int i=1;i<n;i++) 44 { 45 scanf("%d%d",&pa,&val[i]); 46 addedge(pa,i); 47 } 48 int m,pos; 49 scanf("%d",&m); 50 for(int i=0;i<m;i++) 51 { 52 k=0; 53 memset(cnt,0,sizeof(cnt)); 54 scanf("%d",&pos); 55 cnt[k++]=val[pos]; 56 dfs(pos); 57 if(k<3) 58 printf("-1 "); 59 else 60 { 61 sort(cnt,cnt+k); 62 reverse(cnt,cnt+k); 63 printf("%d %d %d ",cnt[0],cnt[1],cnt[2]); 64 } 65 } 66 } 67 68 return 0; 69 }