NOJ [1186] Get the Width

• 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.
  
  
  其实，方法很多，这里我用的是搜索

  //一个结构体来存点，成员有当前rank，左二子，右儿子
  //输入完以后，sort排序，从根节点开始，依次去存好每个节点的等级，
  
  #include<cstdio>
  #include<algorithm>
  #include<cstring>
  using namespace std;
  struct Node
  {
    int left,right;
  }node[25];
  int rank[25];
  bool lp[25],rp[25];
  int lay,alllay;
  void search_rank(int p)
  {
    if(node[p].left!=-1 && lp[p])  //有左儿子
    {
      rank[lay]++;
      int l=node[p].left;
  	lp[p]=false;
  	lay++;
  	alllay++;
  	//printf("递归%d的左子树节点%d ,当前第%d层有节点%d个\n",p,l,lay-1,rank[lay-1]);
      search_rank(l);
    }
    if(node[p].right!=-1 && rp[p])
    {
  
      rank[lay]++;
  	int r=node[p].right;
  	rp[p]=false;
  	lay++;
  	alllay++;
  	//printf("递归%d的右子树节点%d ,当前第%d层有节点%d个\n",p,r,lay-1,rank[lay-1]);
      search_rank(r);
    }
    lay--;
    return ;
  }
  int cmp(const int a,const int b)
  {
    return a>b;
  }
  int main()
  {
    int T,n,i,j;
    int p,pl,pr;
    while(~scanf("%d",&T))
    {
      while(T--)
      {
        scanf("%d",&n);
        memset(rank,0,sizeof(rank));
        memset(lp,true,sizeof(lp));
        memset(rp,true,sizeof(rp));
        for(i=1;i<=n;i++)
        {
          scanf("%d%d%d",&p,&pl,&pr);
          node[p].left=pl;
          node[p].right=pr;
        }
        rank[1]=1;
        alllay=1;
        lay=2;
        search_rank(1);
        sort(rank+1,rank+alllay+1,cmp);
          printf("%d\n",rank[1]);
      }
    }
    return 0;
  }