Fibonacci Tree

Fibonacci Tree

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 13   Accepted Submission(s) : 6

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[sup]5[/sup]) and M(0 <= M <= 10[sup]5[/sup]). 　　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

 

Source

最小生成树和最大生成树中白边的数量之间是否夹了一个菲波数，先打表列出能用到的菲波数，然后最大生成树与最小生成树的建立。代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 1000010
using namespace std;
int sum,ans;
int pri[MAXN];
int fibonacci[26]={0,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025};//打表
struct s
{
	int a;
	int b;
	int cost;
}dis[MAXN];
bool cmp(s A,s B)
{
	return A.cost<B.cost;
}
int find(int x)
{
	int r=x;
	while(r!=pri[r])
	r=pri[r];
	int i=x,j;
	while(i!=r)
	{
		j=pri[i];
		pri[i]=r;
		i=j;
	}
	return r;
}
void connect(int xx,int yy,int num)
{
	int nx=find(xx);
	int ny=find(yy);
	if(nx!=ny)
	{
		pri[nx]=ny;
		ans++;
		if(num==1)
		sum++;
	}
}
int main()
{
	int i,j,m,s,n;
	int t,q,num,k,low,high;
	int cas=0;
	
	scanf("%d",&t);
	while(t--)
	{
		sum=0;
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
		pri[i]=i;
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&dis[i].a,&dis[i].b,&dis[i].cost);
		}
		sort(dis,dis+m,cmp);
		
		ans=0;
		for(i=0;i<m;i++)
		{
			connect(dis[i].a,dis[i].b,dis[i].cost);
		}
		if(ans!=n-1)   
        {  
            printf("Case #%d: No\n",++cas);  
            continue;  
        }
		low=sum;
		
		sum=0;
		for(i=1;i<=n;i++)
		pri[i]=i;
		for(i=m-1;i>=0;i--)
		{
			connect(dis[i].a,dis[i].b,dis[i].cost);
		}
		high=sum;
		
		int bz=0;
		for(i=1;i<=24;i++)
		{
			if(fibonacci[i]>=low&&fibonacci[i]<=high)
			bz=1;
		}
		
		if(bz)
		printf("Case #%d: Yes\n",++cas);
		else
		printf("Case #%d: No\n",++cas);
	}
	return 0;
}