hdu 4786 Fibonacci Tree（最小生成树）

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2952    Accepted Submission(s): 947

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，问连通全部点时的边中，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++)//由于边最多有100000条，所以斐波那契数大于这个数时就能够了，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
");//假设没有连通全部点直接输出No 
		else{
			for(i=1;i<50;i++){//注意这个i要从1開始，由于输入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
");
			else printf("No
");
		}
	} 
	return 0;
}