标签：ini clu using ref limit children its lin tor
题目链接：https://vjudge.net/problem/CodeForces-913B
题目描述：
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
4
1
1
1
Yes
7
1
1
1
2
2
2
No
8
1
1
1
1
3
3
3
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:
大概题意就是给你一颗有根数，如果每个中间结点（或者根结点）的叶的数量大于等于3就是云杉，否则不是。
我们可以递归的进行判断：
1.某结点是否有3个以上的叶结点
2.某结点的非叶结点是否有3个以上的叶结点
#include<set> #include<map> #include<stack> #include<queue> #include<cmath> #include<stdio.h> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl ‘\n‘ #define max(a, b) (a > b ? a : b) #define min(a, b) (a < b ? a : b) #define mst(a) memset(a, 0, sizeof(a)) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; const double pi = acos(-1.0); const double eps = 1e-7; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const int NIL = -1; const int maxn = 1e3+10; struct node { int p, l, r; }; vector<int> trees[maxn]; bool solve(int root) { int cnt = 0; for (int i = 0; i<trees[root].size(); ++i) if (trees[trees[root][i]].empty()) ++cnt; if (cnt < 3) return false; else { for (int i = 0; i<trees[root].size(); ++i) if (!trees[trees[root][i]].empty()) //solve(trees[root][i]); 错误1：更深一步的递归不影响之前的结果 //return solve(trees[root][i]); 错误2：如果第一次递归是true即使后面是false也不影响 if (!solve(trees[root][i])) return false; } return true; } int main(void) { int n; while(~scanf("%d", &n)) { for (int i = 2, node; i<=n; ++i) { scanf("%d", &node); trees[node].push_back(i); } printf(solve(1) ? "Yes\n" : "No\n"); for (int i = 0; i<maxn; ++i) trees[i].clear(); } return 0; }
codeforces - 913B Christmas Spruce（树）
标签：ini clu using ref limit children its lin tor
原文地址：https://www.cnblogs.com/shuitiangong/p/12242271.html
