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HDOJ 题目4786 Fibonacci Tree(克鲁斯卡尔)

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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
 

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ac代码
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
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;i<m;i++)
	{
		int u,v,w;
		u=edge[i].u;
		v=edge[i].v;
		w=edge[i].w;
		int fa=find(u);
		int fb=find(v);
		if(fa!=fb)
		{
			pre[fa]=fb;
			ans+=w;
		}
	}
	int num=0;
	for(i=1;i<=n;i++)
		if(pre[i]==i)
			num++;
		if(num>1)
			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;i<m;i++)
		{
			scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
		}
		qsort(edge,m,sizeof(edge[0]),cmp1);
		l=ku();
		qsort(edge,m,sizeof(edge[0]),cmp2);
		r=ku();
		if(l==-1||r==-1)
		{
			printf("Case #%d: No\n",++c);
		}
		else
		{
			int flag=0;
			for(i=1;i<100100;i++)
			{
				if(a[i]>r)
					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);
		}
	}
}


HDOJ 题目4786 Fibonacci Tree(克鲁斯卡尔)

原文:http://blog.csdn.net/yu_ch_sh/article/details/44566893

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