hdu 4786 Fibonacci Tree 生成树 属于需要仔细想的题目，对图必须非常熟。。今天的首A，太堕落了

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2519    Accepted Submission(s): 804

Problem Description

　　Coach Pang is interested in Fibonacci numbers while Uncle Yang wants him to do some research on Spanning Tree. So Coach Pang decides to solve the following problem:
　　Consider a bidirectional graph G with N vertices and M edges. All edges are painted into either white or black. Can we find a Spanning Tree with some positive Fibonacci number of white edges?
(Fibonacci number is defined as 1, 2, 3, 5, 8, ... )

 

Input

　　The first line of the input contains an integer T, the number of test cases.
　　For each test case, the first line contains two integers N(1 <= N <= 10 ⁵) and M(0 <= M <= 10 ⁵).
　　Then M lines follow, each contains three integers u, v (1 <= u,v <= N, u<> v) and c (0 <= c <= 1), indicating an edge between u and v with a color c (1 for white and 0 for black).

 

Output

　　For each test case, output a line “Case #x: s”. x is the case number and s is either “Yes” or “No” (without quotes) representing the answer to the problem.

 

Sample Input

  

   2
4 4
1 2 1
2 3 1
3 4 1
1 4 0
5 6
1 2 1
1 3 1
1 4 1
1 5 1
3 5 1
4 2 1
  

 

Sample Output

  

   Case #1: Yes
Case #2: No
  

 

这道题如果不仔细想的话，，还是挺难的，首先题目的意思是通过图生成一课树，判断生成的树中白色结点的个数恰好是Fibonacci 数，

关于这道题的思路，请看这篇博客： http://blog.youkuaiyun.com/dyx404514/article/details/16371907#cpp

博主的证明的思路不是太通顺，，你看到结论后，应该自己去思考一下是不是这样

代码：

#include <stdio.h>
#define MAX 101000
struct Edge{
	int x , y , c ;
}edge[MAX];
int f[MAX] ;
bool fib[MAX] ;
int find(int x)
{
	int r = x ;
	while(r != f[r])
	{
		r = f[r] ;
	}
	int temp ;
	while(x != r)
	{
		temp = f[x] ;
		f[x] = r ;
		x = temp ;
	}
	return r ;
}

int count(int n , int m , int c)
{
	for(int i = 0 ; i <= n ; ++i)
	{
		f[i] = i ;
	}
	int sum = 0 ;
	for(int i = 0 ; i < m ; ++i)
	{
		if(edge[i].c != c)
		{
			int x = find(edge[i].x) , y = find(edge[i].y) ;
			if(x != y)
			{
				sum ++ ;
				f[x] = f[y] ;
			}
		} 
		
	}
	return sum ;
}

int main()
{
	int t , c = 1;
	int a = 1 , b = 0;
	while(a+b < MAX)
	{
		fib[a+b] = true ;
		int t = b ;
		b = a ;
		a = t+a ;
	}
	scanf("%d",&t);
	while(t--)
	{
		int n , m ;
		scanf("%d%d",&n,&m) ;
		for(int i = 0 ; i < m ; ++i)
		{
			scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].c) ;
		}
		printf("Case #%d: ",c++);
		int sum = count(n,m,2) ;
		if(sum < n-1)
		{
			puts("No") ;
			continue ;
		}
		int max = count(n,m,0);
		int min = n-1-count(n,m,1) ;
		bool flag = false ;
		for(int i = min ; i <= max ; ++i)
		{
			if(fib[i] == true)
			{
				flag = true ;
				break ;
			}
		}
		if(flag)
		{
			puts("Yes");
		}
		else
		{
			puts("No") ;
		}
	}
	return 0 ;
}

与君共勉