Hdu4786

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2340    Accepted Submission(s): 748

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");

        }

    }
}