HDOJ 4786 Fibonacci Tree（最小生成树--kruskul）

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3004    Accepted Submission(s): 964

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 ⁵) and M(0 <= M <= 10 ⁵).
　　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


    
    
    
     看下数据范围，果断kruskul，这道题的意思是，看一下是否建立一个连通图，然后 其白色边的数量等于一个fibonacci数，因为数据范围最大为100000，所以直接打 表fibonacci数，然后求得最大生成树白色边数量和最小生成树白色边数量，然后查看 是否有fibonacci数在这个范围中即可

ac代码：

    
    
    
    
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 1000010
#define INF 0xfffffff
#define min(a,b) (a>b?b:a)
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++;//记录节点数量
		if(num==1)//记录白色边数量
		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++)//最小生成树白色边数量
		{
			connect(dis[i].a,dis[i].b,dis[i].cost);
		}
		if(ans!=n-1)  //如果不能建立连通图，那么不用执行后面的了，直接输出No
        {  
            printf("Case #%d: No\n",++cas);  
            continue;  
        }
		low=sum;
		sum=0;
		for(i=1;i<=n;i++)//初始化联通
		pri[i]=i;
		for(i=m-1;i>=0;i--)//最大生成树白色边数量
		{
			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;
}