Christmas Spruce
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called a child 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.
Input
The 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.
Output
Print “Yes” if the tree is a spruce and “No” otherwise.
Examples
input
4
1
1
1
output
Yes
input
7
1
1
1
2
2
2
output
No
input
8
1
1
1
1
3
3
3
output
Yes
Note
The first example:
The second example:
It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.
The third example:
打cf从来没想过涨分!只求不掉分就是我最大的愿望了,这场hello 2018是当时在学校写的..ac a和b后就溜去睡觉了((つД`)太弱了…),第二天起来一看被hack了(;д;)…掉了一大波分..不想缩话
题意:
英语不好,看图理解..看给的这个树的非叶节点上是不是都有三个或以上的节点,是的话就是一棵云杉什么的(云杉是百度翻译??),是就输出yes,否则no。
想法:
纯模拟,开一个结构体记录双亲是谁与儿子的数量,然后就模拟咯..
以下是被hack后改正确的代码…o(╥﹏╥)o
#include <iostream>
#include <cstring>
using namespace std;
struct tree
{
int parent;
int son;
};
tree T[1005];
int main()
{
int n;
int i;
int v[1005];
memset(v, 0, sizeof(v));
for(i = 0; i < 1005; i++)
{
T[i].parent = 0;
T[i].son = 0;
}
cin >> n;
int x;
for(i = 2; i <= n; i++)
{
cin >> x;
if(!v[x])
{
T[T[x].parent].son--;
v[x] = 1;
}
T[i].parent = x;
T[x].son++;
}
int flag = 0;
for(i = 1; i <= n; i++)
{
if((T[i].son && T[i].son < 3)||(v[i] && T[i].son < 3))
{
flag = 1;
break;
}
}
if(flag)
{
cout << "No" << endl;
}
else
{
cout << "Yes" << endl;
}
return 0;
}