Fibonacci Tree 4786

Fibonacci Tree

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

Total Submission(s) : 6   Accepted Submission(s) : 3

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
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int a[100010];
int fb[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};//数很少，直接打表 
int sum,ans;
struct zz
{
	int f1,f2,c;
}q[100010];
int cmp(zz x,zz y)
{
	return x.c<y.c;
}
/*int find(int x)  //这种超时 
{
	while(x!=a[x])
		x=a[x];
		return x;
}*/
int find (int x)
{
	int r=x;
	while(r!=a[r])
		r=a[r];
	int i=x,j;
	while(i!=r)
	{
		j=a[i];
		a[i]=r;
		i=j;
	}
	return r;
}
void marge(int x,int y,int z)
{
	int fx,fy;
	fx=find(x);
	fy=find(y);
	if(fx!=fy)
	{
		a[fx]=fy;
		ans++;	//记录节点数，方便后面判断是否能形成树。 
		if(z==1)
			sum++;//记录白边的个数 
	}
}
int main()
{
	int i,j,m,n,zd,zs,flag;
	int t,s=0;
	scanf("%d",&t);
	while(t--)
	{
		sum=0;ans=0;
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
			a[i]=i;
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&q[i].f1,&q[i].f2,&q[i].c);
		}
		sort(q,q+m,cmp);//最小生成树，求出白边最少有几条 
		for(i=0;i<m;i++)
		{
			marge(q[i].f1,q[i].f2,q[i].c);
		}
		if(ans!=n-1)//如果不能形成树，直接输出No。 
		{
			printf("Case #%d: No\n",++s);
			continue;
		}
		zs=sum;//最少白边数量。 
		
		sum=0;//第二步，求最多有几条白边 
		for(i=1;i<=n;i++)
			a[i]=i;
		for(i=m-1;i>=0;i--)
		{
			marge(q[i].f1,q[i].f2,q[i].c);
		}
		zd=sum;//最多白边数量 
		flag=0;
		for(i=1;i<25;i++)
		{
			if(fb[i]>=zs&&fb[i]<=zd)
				flag=1;
		}
		if(flag)
			printf("Case #%d: Yes\n",++s);
		else
			printf("Case #%d: No\n",++s);
	}
	return 0;
}