hdu4786-图论训练3-最小生成树

D - Fibonacci Tree

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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 
      
      
      
      

 题意：给你n个顶点m条边，边有黑白两种颜色，用这些边来建一个生成树（可以使用任意边，黑边白边都可以），在所有的生成树中使用白边的数目形成一个数列，问这个数列中有没有斐波那契数列里数字；

  我们可以求出用白边最少和用白边最多的两课生成树，然后得到两个数字，判断这两个数字之间有没有斐波那契数即可，

我的代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=400000;
int fib[60];
int n,m;
struct Edge
{
    int u,v,w;
}d[maxn];
bool cmp1(Edge a,Edge b)
{
    return a.w<b.w;
}
bool cmp2(Edge a,Edge b)
{
    return a.w>b.w;
}

void get_fib()
{
    fib[0]=0;
    fib[1]=1;
    int i;
    for(i=2;i<=26;i++)
        fib[i]=fib[i-1]+fib[i-2];

}
int par[maxn],rak[maxn];
void init()
{
    for(int i=0;i<=n;i++)
    {
        par[i]=i;
        rak[i]=0;
    }
}
int find_(int x)
{
    if(par[x]==x)
        return x;
    else
        return par[x]=find_(par[x]);
}
void unite(int x,int y)
{
    x=find_(x);
    y=find_(y);
    if(x==y)
        return;
    if(rak[x]<rak[y])
    {
        par[x]=y;
    }
    else
    {
        par[y]=x;
        if(rak[x]==rak[y])
            rak[x]++;
    }
}
bool vis[maxn];
int main()
{
    int t,T=0;
    int l,r;
    get_fib();
    cin>>t;
    while(t--)
    {
        scanf("%d%d",&n,&m);
        l=0,r=0;
        for(int i=0;i<m;i++)
        scanf("%d%d%d",&d[i].u,&d[i].v,&d[i].w);

        printf("Case #%d: ",++T);

        init();///白边为辅
        sort(d,d+m,cmp1);
        for(int i=0;i<m;i++)
        {
            Edge e=d[i];
            if(find_(e.u)!=find_(e.v))
            {

                unite(e.u,e.v);
                l+=e.w;
            }
        }

        init();///白边为主
        memset(vis,false,sizeof(vis));
        sort(d,d+m,cmp2);
        for(int i=0;i<m;i++)
        {
            Edge e=d[i];
            if(find_(e.u)!=find_(e.v))
            {
                vis[e.u]=vis[e.v]=true;
                unite(e.u,e.v);
                r+=e.w;
            }
        }
        
        bool fg=true;
         for(int i=1;i<=n;i++)///检测是否所有顶点入集合，保证是生成树啊
         {
             if(!vis[i])
                {
                    fg=false;
                    break;
                }
         }
         if(!fg)
         {
             cout<<"No"<<endl;
             continue;
         }
         
        int flag=0;
        for(int i=1;i<=26;i++)
        {
            if(l<=fib[i]&&r>=fib[i])
            {
                flag=1;
            }
        }
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");

    }
    return 0;
}