2013 ICPC亚洲区域赛成都站 F Fibonacci Tree（最小生成树）

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

题意：

有一个n个点，m条边的图，给定边的权值为1（白色）或2（黑色），问是否存在一个生成树，使得其中白边的数量为斐波那契数？

题解：

首先判断这个图是否为连通图，若不是直接输出No。

然后只要用白边优先（最大生成树）的总权值减去黑边优先（最小生成树）的总权值，就可以得到一个白边数量的区间，然后枚举斐波那契数即可。

题目链接：HDOJ 4786

#include<bits/stdc++.h>
#define MAX 100000
using namespace std;
int n,m,p1[MAX+5],p2[MAX+5],fib[55];
struct edge{
    int u,v,w;
}e[MAX+5];
int find(int r,int p[])
{
    if(p[r]!=r)
        p[r]=find(p[r],p);
    return p[r];
}
void init(int p[])
{
    for(int i=0;i<=MAX;i++)
        p[i]=i; 
}
bool cmp1(edge a,edge b){return a.w>b.w;}
bool cmp2(edge a,edge b){return a.w<b.w;}
int KurskalMax(int p[])
{
    init(p);
    sort(e,e+m,cmp1);
    int cnt=0,cost=0,i;
    for(i=0;i<m;i++)
    {
        int fu=find(e[i].u,p),fv=find(e[i].v,p);
        if(fu!=fv)
        {
            p[fu]=fv;
            cost+=e[i].w;
            cnt++;
        }
        if(cnt==n-1)break;
    }
    return cost;
}
int KurskalMin(int p[])
{
    init(p);
    sort(e,e+m,cmp2);
    int cnt=0,cost=0,i;
    for(i=0;i<m;i++)
    {
        int fu=find(e[i].u,p),fv=find(e[i].v,p);
        if(fu!=fv)
        {
            p[fu]=fv;
            cost+=e[i].w;
            cnt++;
        } 
        if(cnt==n-1)break;
    }
    if(cnt!=n-1)return -1;
    return cost;
}
int main()
{
    int i;
    fib[1]=1;fib[2]=2;
    for(int i=3;i<=40;i++)
        fib[i]=fib[i-1]+fib[i-2];
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%d%d",&n,&m); 
        for(i=0;i<m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            e[i].u=u;e[i].v=v;e[i].w=w;
        }
        int L=KurskalMin(p1),R=KurskalMax(p2),flag=0;
        if(L!=-1)
        {
            for(i=L;i<=R;i++)
            {
                if(fib[lower_bound(fib+1,fib+40+1,i)-fib]==i)
                {
                    flag=1;
                    break;
                }
            }
        }
        printf("Case #%d: ",cas);
        if(flag)printf("Yes
");
        else printf("No
");
    } 
    return 0;
}