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)

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

#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;
}