Hello 2018 B. Christmas Spruce 水题

B. Christmas Sprucetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputConsider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called achild of vertex v and vertex v is called a parent of vertex u if there exists a directed edge from v to u. A vertex is called a leaf if it doesn't have children and has a parent.Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.The definition of a rooted tree can be found here.InputThe first line contains one integer n — the number of vertices in the tree (3 ≤ n ≤ 1 000). Each of the next n - 1 lines contains one integer pi (1 ≤ i ≤ n - 1) — the index of the parent of the i + 1-th vertex (1 ≤ pi ≤ i).Vertex 1 is the root. It's guaranteed that the root has at least 2 children.OutputPrint "Yes" if the tree is a spruce and "No" otherwise.ExamplesinputCopy

4
1
1
1

output

Yes

inputCopy

7
1
1
1
2
2
2

output

No

inputCopy

8
1
1
1
1
3
3
3

output

Yes

NoteThe first example:

The second example:

It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.The third example:

题意：构造一棵树，判断其每一个非叶子结点时候有3个以上叶子结点。
代码如下：#include <bits/stdc++.h>
using namespace std;
int G[1100][1100];
int leaf[1100];
int main()
{
int n;
while (~scanf("%d",&n)){
memset(G,0,sizeof(G));
memset(leaf,0,sizeof(leaf));
int u;
for (int v=2;v<=n;v++){
scanf("%d",&u);
G[u][v]=1;
leaf[u]=1;
}
int f=0;
for (int i=1;i<=n;i++){
int cnt=0;
if (!leaf[i])
continue;
for (int j=1;j<=n;j++){
if (G[i][j]&&!leaf[j])
cnt++;
}
if (cnt<3){
f=1;
break;
}
}
if (f)
printf("no\n");
else
printf("yes\n");
}
return 0;
}