Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

有n个点，m条边，有黑白之分，问连通所有点时的边中，白边的个数能不能是斐波那契数列中的一个数

思路：先用白边连图，求出白边的个数，再先用黑边连图，求出白边另个值，，这两个值就是白边个数的取值范围

2015,7,30

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define M 100000+10
struct node{
	int s,e,val;
}sd[M];
int x[M];
int a[55];

bool cmp1(node a,node b){
	return a.val>b.val;
}
bool cmp2(node a,node b){
	return a.val<b.val;
}
void init()
{
	for(int i=0;i<M;i++)
		x[i]=i;
}
int find(int k)
{
	if(x[k]==k) return k;
	x[k]=find(x[k]);
	return x[k];
}
int main()
{
	int t,m,n,v=1,i;
	int num,start,end;

	a[1]=1; a[2]=2;
	for(i=3;i<55;i++)//因为边最多有100000条，所以斐波那契数大于这个数时就可以了，55足够了
		a[i]=a[i-1]+a[i-2];

	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&m);
		for(i=0;i<m;i++)
			scanf("%d%d%d",&sd[i].s,&sd[i].e,&sd[i].val);

		sort(sd,sd+m,cmp2);
		num=0; 	init();
		for(i=0;i<m;i++){
			int fa=find(sd[i].s);
			int fb=find(sd[i].e);
			if(fa!=fb){
				x[fa]=fb;
				if(sd[i].val==1)
					num++;
			}
		}
		start=num;

		sort(sd,sd+m,cmp1);
		num=0;	init();
		for(i=0;i<m;i++){
			int fa=find(sd[i].s);
			int fb=find(sd[i].e);
			if(fa!=fb){
				x[fa]=fb;
				if(sd[i].val==1)
					num++;
			}
		}
		end=num;

		int ok=0;
		for(i=1;i<=n;i++){
			if(find(i)!=find(1)){
				ok=1;
				break;
			}
		}
		printf("Case #%d: ",v++);
		if(ok) printf("No\n");//如果没有连通所有点直接输出No
		else{
			for(i=1;i<50;i++){//注意这个i要从1开始，因为输入1个点0条边时要输出No，，我找了半天错= =+
				if(a[i]>= start && a[i]<=end){
					ok=1;
					break;
				}
			}
			if(ok) printf("Yes\n");
			else printf("No\n");
		}
	}
	return 0;
}