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

题意：问构成的生成树当中是否存在黑色边（边为1）数为斐波那契数

思路：求出生成树中最小包括的黑色边数。和最多黑色边数，假设有斐波那契数在两者之间，则能够构成。由于黑白边能够搭配使用

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int n,m;
int fibo[50];
int f[100010];
struct node
{
    int u,v,c;
} s[100010];

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

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

int find(int x)
{
    return x == f[x] ? x : f[x] = find(f[x]);
}

void Union(int x ,int y)
{
    int fx = find(x);
    int fy = find(y);

    if(fx != fy)
    {
        f[fx] = fy;
    }
}

int main()
{
#ifdef xxz
    freopen("in.txt","r",stdin);
#endif
    fibo[1] = 1;
    fibo[2] = 2;
    for(int i = 3; ; i++)
    {
        fibo[i] = fibo[i-1] + fibo[i-2];
        if(fibo[i] >= 100000) break;
    }

    int T,Case = 1;;
    scanf("%d",&T);

    while(T--)
    {

        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++) f[i] = i;

        for(int i = 0; i < m; i++)
        {

            scanf("%d%d%d",&s[i].u,&s[i].v,&s[i].c);
            Union(s[i].u,s[i].v);
        }
        int cent = 0;
        int bl = 0, bh = 0;
        int root = 0,size = 0;

        for(int i = 1; i <= n; i++)
        {
            if(f[i] == i)
            {
                cent++;
                root = i;
            }
        }

        printf("Case #%d: ",Case++);
        if(cent >= 2) cout<<"No"<<endl;//首先要推断能否构成一个生成树。推断根节点个数是否为1即可
        else
        {
            sort(s,s+m,cmp1);
            for(int i = 1; i <= n; i++) f[i] = i;

            for(int i = 0; i < m; i++)
            {
                int fu = find(s[i].u);
                int fv = find(s[i].v);
                if(fu == fv) continue;

                bl += s[i].c;
                size++;
                Union(s[i].u,s[i].v);
                if(size == n-1) break;
            }

            size = 0;
            sort(s,s+m,cmp2);
            for(int i = 1; i <= n; i++) f[i] = i;

            for(int i = 0; i < m; i++)
            {
                int fu = find(s[i].u);
                int fv = find(s[i].v);
                if(fu == fv) continue;

                bh += s[i].c;
                size++;
                Union(s[i].u,s[i].v);
                if(size == n-1) break;
            }

            int flag = 0;
            for(int i =1; fibo[i] <= 100000 ; i++ )
            {
                if(fibo[i] >= bl && fibo[i] <= bh)
                {
                    flag = 1;
                    break;
                }
            }
            if(flag) printf("Yes\n");
            else printf("No\n");

        }

    }
}