BestCoder Round #53 (div.2) HDOJ5423 Rikka with Tree(bfs)

Rikka with Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 312 Accepted Submission(s): 156



Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

For a tree T,
let F(T,i) be
the distance between vertice 1 and vertice i.(The
length of each edge is 1).

Two trees A and B are
similiar if and only if the have same number of vertices and for each i meet F(A,i)=F(B,i).

Two trees A and B are
different if and only if they have different numbers of vertices or there exist an number i which
vertice i have
different fathers in tree A and
treeB when
vertice 1 is root.

Tree A is
special if and only if there doesn't exist an tree B which A and B are
different and A and B are
similiar.

Now he wants to know if a tree is special.

It is too difficult for Rikka. Can you help her?



Input

There are no more than 100 testcases.

For each testcase, the first line contains a number n(1≤n≤1000).

Then n−1 lines
follow. Each line contains two numbers u,v(1≤u,v≤n) ,
which means there is an edge between u and v.



Output

For each testcase, if the tree is special print "YES" , otherwise print "NO".



Sample Input

3
1 2
2 3
4
1 2
2 3
1 4

Sample Output

YES
NO

HintFor the second testcase, this tree is similiar with the given tree:
4
1 2
1 4
3 4

问你一棵树是否特殊，一棵树特树当且仅当非叶子结点个数不多于一个。 bfs写一遍～

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "vector"
#include "queue"
using namespace std;
const int MAXN = 1005;
int n;
bool vis[MAXN];
vector<int> v[MAXN];
void bfs()
{
	queue<int> q;
	int sum = 1;
	q.push(1);
	vis[1] = true;
	while(!q.empty()) {
		int x = q.front(), tmp = 0;
		q.pop(); 
		for(int i = 0; i < v[x].size(); ++i) {
			int y = v[x][i];
			if(vis[y]) continue;
			vis[y] = true;
			tmp++;
			q.push(y);
		}
		sum += tmp;
		if(tmp != 1) break;
	}
	if(sum == n) printf("YES\n");
	else printf("NO\n");
}
int main(int argc, char const *argv[])
{
	while(scanf("%d", &n) != EOF) {
		memset(vis, false, sizeof(vis));
		for(int i = 0; i <= n; ++i)
			v[i].clear();
		for(int i = 0; i < n - 1; ++i) {
			int x, y;
			scanf("%d%d", &x, &y);
			v[x].push_back(y);
			v[y].push_back(x);
		}
		bfs();
	}
	return 0;
}