http://cdn.ac.nbutoj.com/Problem/view.xhtml?id=1186

• 问题描述
• It's an easy problem. I will give you a binary tree. You just need to tell me the width of the binary tree.
  The width of a binary tree means the maximum number of nodes in a same level.
  
  

  

  
  For example, the width of binary tree above is 3.
  

  
  
• 输入
• The first line is an integer T, means the number of cases.
  Then follow T cases.
  For each case, the first line is an integer N, means the number of nodes. (1 <= N <= 10)
  Then follow N lines. Each line contains 3 integers P A B; indicate the number of this node and its two children node. If the node doesn’t have left child or right child, then replace it by -1.
  You can assume the root is 1.
  
• 输出
• For each case, output the width.
• 样例输入

• 1
  6
  4 -1 -1
  2 4 5
  5 -1 -1
  1 2 3
  6 -1 -1
  3 -1 6
  

• 样例输出

• 3

  AC代码：

  #include<iostream>
  #include<string.h>
  #include<string>
  #include<algorithm>
  #include<cstdio>
  #define N 20
  using namespace std;
  typedef struct 
  {
  	int to;
  	int next;
  }Node;
  Node s[3*N];
  int lev[N];
  int head[N];
  int res;
  void init()
  {
  	res=0;
  	memset(lev,0,sizeof(lev));
  	memset(head,-1,sizeof(head));
  
   }
  void add(int a,int b)
  {
  	s[res].to=b;
  	s[res].next=head[a];
  	head[a]=res++;
  }
  void dfs(int cur,int now)
  {
  	 lev[cur]++;
  	 if(head[now]==-1) return;
  	 for(int i=head[now];i!=-1;i=s[i].next)  dfs(cur+1,s[i].to);
  }
  int main()
  {
  	int T;
      scanf("%d",&T);
  	while(T--)
  	{
  		int n;
  		init();
  		scanf("%d",&n);
  		for(int i=0;i!=n;++i)
  		{
  			int a,b,c;
  			scanf("%d%d%d",&a,&b,&c);
  			if(b!=-1) add(a,b);
  			if(c!=-1) add(a,c);
  		}
  		dfs(0,1);
  		printf("%d\n",*max_element(lev,lev+n));
  	}return 0;
  }