hdu 2044 一只小蜜蜂 题解

一只小蜜蜂...

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 63088    Accepted Submission(s): 22872

Problem Description

有一只经过训练的蜜蜂只能爬向右侧相邻的蜂房，不能反向爬行。请编程计算蜜蜂从蜂房a爬到蜂房b的可能路线数。
其中，蜂房的结构如下所示。

 

Input

输入数据的第一行是一个整数N,表示测试实例的个数，然后是N 行数据，每行包含两个整数a和b(0<a<b<50)。

 

Output

对于每个测试实例，请输出蜜蜂从蜂房a爬到蜂房b的可能路线数，每个实例的输出占一行。

 

Sample Input

  

   2
1 2
3 6
  

 

Sample Output

  

   1
3
  

 

解题思路：

如果两点的差为n，若n为0或1，那么路径数为1，如果n超过1，则r(n)=r(n-1)+r(n-2)

代码展示：

<span style="font-size:14px;">#include <iostream>
#include <cstring>
using namespace std;
long long ans[50]={0};
long long cal(int l)
{
	if(l==0||l==1) return 1;
	if(!ans[l]) return ans[l]=cal(l-1)+cal(l-2);
	return ans[l];
}
int main()
{
	int n;
	cin>>n;
	while(n--)
	{
		int l1 , l2;
		cin>>l1>>l2;
		int l = l2-l1;
		memset(ans,0,50);
		cout<<cal(l)<<endl;
	}
}</span>

310. Minimum Height Trees

 

Question Editorial Solution

  My Submissions

• Total Accepted: 21692
• Total Submissions: 77582
• Difficulty: Medium
• Contributors: Admin

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
      \ | /
        3
        |
        4
        |
        5

return [3, 4]


解题思路：

本题即要找到所有的满足以该结点为根的树的高度最小的节点。由树的性质（连通的无环图）可以得到这样的节点最多只有两个。如果有三个则不满足无环额度条件。所以当结点总数大于2时，算法的每一步可以将最外层的结点删掉，以为以最外层的结点为根的子树高度必然要大一些。当剩余节点总数小于2时，即将其加入结果向量。

另外一种朴素的想法就是利用宽度优先搜索，分别确定以每一个结点为根节点的树的高度，然后再挑选出树的高度最小的。显然这种方法的复杂度相对于第一种来说要高，在本题使用这种算法会导致超时。

代码展示：

class Solution {
public:


vector<int> findMinHeightTrees(int n, vector<pair<int, int> >& edges)
{
<span style="white-space:pre">	</span>vector<int> g[n];
<span style="white-space:pre">	</span>vector<int> ans;
<span style="white-space:pre">	</span>if(n==1) 
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>ans.push_back(0);<span style="white-space:pre">	</span>
<span style="white-space:pre">		</span>return ans;
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>int esize = edges.size();
<span style="white-space:pre">	</span>vector<int>indeg(n);
<span style="white-space:pre">	</span>//for(int i=0;i<n;i++) indeg[i]=0;
<span style="white-space:pre">	</span>for(int i=0;i<esize;i++)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>pair<int,int> tmp = edges[i];
<span style="white-space:pre">		</span>g[tmp.first].push_back(tmp.second);
<span style="white-space:pre">		</span>g[tmp.second].push_back(tmp.first);
<span style="white-space:pre">		</span>indeg[tmp.first]++;
<span style="white-space:pre">		</span>indeg[tmp.second]++;
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>queue<int> q;
<span style="white-space:pre">	</span>for(int i=0;i<n;i++)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>if(indeg[i]==1)
<span style="white-space:pre">		</span>q.push(i);
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>while(!q.empty())
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>int size = q.size();
<span style="white-space:pre">		</span>for(int i=0;i<size;i++)
<span style="white-space:pre">		</span>{
<span style="white-space:pre">			</span>int tmp = q.front();
<span style="white-space:pre">			</span>q.pop();
<span style="white-space:pre">			</span>indeg[tmp]--;
<span style="white-space:pre">			</span>if(n<=2) ans.push_back(tmp);
<span style="white-space:pre">			</span>for(int i=0;i<g[tmp].size();i++)
<span style="white-space:pre">			</span>{
<span style="white-space:pre">				</span>indeg[g[tmp][i]]--;
<span style="white-space:pre">				</span>if(indeg[g[tmp][i]]==1) q.push(g[tmp][i]);
<span style="white-space:pre">			</span>}
<span style="white-space:pre">		</span>}
<span style="white-space:pre">		</span>n-=size;
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>return ans;
<span style="white-space:pre">	</span>
}
};