Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2562 Accepted Submission(s): 816


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 <= 10⁵) and M(0 <= M <= 10⁵).
　　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 2013 Asia Chengdu Regional Contest
Recommend We have carefully selected several similar problems for you: 5193 5192 5191 5190 5189 ac代碼

#include
#include
#include
struct s
{
	int u,v,w;
}edge[100100];
int pre[100100],a[100100],n,m;
int cmp1(const void *a,const void *b)
{
	return (*(struct s *)a).w-(*(struct s *)b).w;
}
int cmp2(const void *a,const void *b)
{
	return (*(struct s *)b).w-(*(struct s *)a).w;
}
void init(int n)
{
	int i;
	for(i=0;i<=n;i++)
		pre[i]=i;
}
void fun()
{
	int i;
	a[1]=1;a[2]=2;
	for(i=3;i<100100;i++)
		a[i]=a[i-1]+a[i-2];
}
int find(int x)
{
	if(x==pre[x])
		return pre[x];
	return pre[x]=find(pre[x]);
}
int ku()
{
	init(n);
	int ans=0,i,j;
	for(i=0;i1)
			return -1;
		return ans;
}
int main()
{
	int t,c=0;
	scanf("%d",&t);
	fun();
	while(t--)
	{
		//int n,m
		int i,l,r;
		scanf("%d%d",&n,&m);
		//init(n);
		for(i=0;ir)
					break;
				if(a[i]<=r&&a[i]>=l)
				{
					printf("Case #%d: Yes\n",++c);
					flag=1;
					break;
				}
			}
			if(!flag)
				printf("Case #%d: No\n",++c);
		}
	}
}