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

这题要求生成树的 边权和 为 斐波那契值；

一个生成树构造成另一个生成树，可以不断的去边又加边来完成。 而这题全部边权为01，所以加边去边过程是+1 +1 的；

可以求出生成树 最大边权和 ，及最小的边权和。  然后枚举24个 小于100000的斐波那契数。

只要有某个斐波那契数  fb，    fb<=最大边权和&&fb>=最小边权和  那就是可以构成这样一颗树的；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f[100100];
int tt[25];
int find(int x){return x==f[x]?x:f[x] = find(f[x]);}  //初始化为自己
void Union(int x, int y){
    int fx = find(x), fy = find(y);
    if(fx == fy)return ;
    if(fx>fy) swap(fx,fy);
    f[fx] = f[x] = f[y] = fy;//这样 就可以直接查f[i]  判断祖先了
}
struct path
{
    int v,u,c;
};
path lu[200100];

int cmp(path a,path b)// 白先
{
    return a.c>b.c;
}

int cmp2(path a,path b)
{
    return a.c<b.c;
}
int main()
{
    int i,t;
    int big,sml;
    tt[1]=1,tt[2]=2;
    for(i=3;;i++)
    {
        tt[i]=tt[i-2]+tt[i-1];
        if(tt[i]>100000)
            break;
    }
    int cas=1;
    int u,v,n,m,flag;
    int x,y;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&lu[i].u,&lu[i].v,&lu[i].c);
        }

        flag=1;
        // 求大的
        for(i=1;i<=n;i++)
            f[i]=i;
        sort(lu,lu+m,cmp);
        big=0;
        for(i=0;i<m;i++)
        {
            x=lu[i].v;
            y=lu[i].u;
            if(find(x)!=find(y))
            {
                Union(x,y);
                big+=lu[i].c;
            }
        } 

        for(i=1;i<=n;i++)//
        {
            if(find(i)!=find(1))
                flag=0;
        }

        sml=0;
        for(i=1;i<=n;i++)
            f[i]=i;
        sort(lu,lu+m,cmp2);
        for(i=0;i<m;i++)
        {
            x=lu[i].v;
            y=lu[i].u;
            if(find(x)!=find(y))
            {
                Union(x,y);
                sml+=lu[i].c;
            }
        } 

        for(i=1;i<=n;i++)//
        {
            if(find(i)!=find(1))
                flag=0;
        }

        printf("Case #%d: ",cas++);
        if(flag==0)
        {
            printf("No\n");
            continue;
        }

    //    printf("%d\n",tt[25]);
        flag=0;
        for(i=1;i<=24;i++)
        {
            if(tt[i]<=big&&tt[i]>=sml)
                flag=1;
        }
        if(flag)
            puts("Yes");
        else
            puts("No");
    }
    return 0;
}