HDOJ 4786 Fibonacci Tree

最大生成树夹最小生成树，老题目了，依稀记得当年在成都靠这题捡了个铜。。。。。

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1572    Accepted Submission(s): 479

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
  
  

 

Source

2013 Asia Chengdu Regional Contest

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=100100;

int nf,fib[100];

int getFib()
{
    fib[0]=1;fib[1]=2;
    nf=2;
    for(int i=2;fib[nf-1]<=100100;i++)
    {
        fib[nf]=fib[nf-1]+fib[nf-2];
        nf++;
    }
}

int n,m;
int fa[maxn];

int find(int x)
{
    if(x==fa[x]) return x;
    return fa[x]=find(fa[x]);
}

struct Edge
{
    int u,v,c;
}edge[maxn];

bool cmp1(Edge x,Edge y)
{
    return x.c<y.c;
}

bool cmp2(Edge x,Edge y)
{
    return x.c>y.c;
}

int Kruscal()
{
    int cnt=n,ans=0;
    for(int i=0;i<=n+1;i++) fa[i]=i;
    for(int i=0;i<m;i++)
    {
        int f1=find(edge[i].u);
        int f2=find(edge[i].v);

        if(f1!=f2)
        {
            fa[f1]=f2;
            ans+=edge[i].c;
            cnt--;
            if(cnt==1) break;
        }
    }
    return (cnt==1)?ans:0x3f3f3f3f;
}

int main()
{
    getFib();
    int T_T,cas=1;
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d%d",&n,&m);

        for(int i=0;i<m;i++)
        {
            int a,b,w;
            scanf("%d%d%d",&a,&b,&w);
            edge[i].u=a; edge[i].v=b; edge[i].c=w;
        }
        sort(edge,edge+m,cmp1);
        int MiMST=Kruscal();
        sort(edge,edge+m,cmp2);
        int MxMST=Kruscal();

        bool flag=false;
        for(int i=0;i<nf;i++)
        {
            if(fib[i]>=MiMST&&fib[i]<=MxMST)
            {
                flag=true; break;
            }
        }

        printf("Case #%d: ",cas++);
        if(flag) puts("Yes");
        else puts("No");
    }
    return 0;
}