題意:一共n個點m條邊，每條邊有個權值1代表白邊，0代表黑邊，求能否建一個生成樹，使得白邊的數量是一個斐波那契數列

思路:求得是白邊的數，可以求出白邊的上限與下限，再在這個區間內找有沒有屬於斐波拉契數列中的數，求上下限時用kruskal演算法，優先選用白邊，求出白邊數的上限，優先選用黑邊，求出白邊數的下限。

#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
 

#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
int a[maxn];
struct node {
    int u, v,
 
 w;
    node() {}
    node(int u, int v, int w): u(u), v(v), w(w) {}
} e[2 * maxn];
int tot;
bool cmp1(node a, node b) {
    return a.w > b.w;
}
bool cmp2(node a, node b) {
    return a.w < b.w;
}
int f[maxn];
int getf(int v) {
    return f[v] == v ? v : f[v] = getf(f[v]);
}

int T, n, m;
int  flag;
 

bool vis[maxn];

int main() {
    a[1] = 1;
    a[2] = 2;
    for (int i = 3; i <= 100000; i++) {
        if (a[i - 1] > 1e5) break;
        a[i] = a[i - 1] + a[i - 2];
    }
    scanf("%d", &T);
    for (int cas = 1; cas <= T; cas++) {
        printf("Case #%d: ", cas);
        tot = 0;
        flag = 0;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) f[i] = i;

        for (int i = 1; i <= m; i++) {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            e[tot++] = {u, v, w};
        }
        sort(e, e + tot, cmp2);
        int MIN = 0;
        int cnt = 0;
        for (int i = 0; i < tot; i++) {
            if (cnt == n - 1) break;
            int t1 = getf(e[i].u);
            int t2 = getf(e[i].v);
            if (t1 != t2) {
                f[t2] = t1;
                cnt++;
                MIN += e[i].w;
            }
        }
        if (cnt != n - 1) {
            printf("No\n");
            continue;
        }
        for (int i = 1; i <= n; i++) f[i] = i;
        int MAX = 0;
        sort(e, e + tot, cmp1);
        cnt = 0;
        for (int i = 0; i < tot; i++) {
            int t1 = getf(e[i].u);
            int t2 = getf(e[i].v);
            if (cnt == n - 1) break;
            if (t1 != t2) {
                f[t2] = t1;
                cnt++;
                MAX += e[i].w;
            }
        }

        for (int i = 1; i <= 100009; i++) {
            if (a[i] > MAX) break;
            if (a[i] <= MAX && a[i] >= MIN) flag = 1;
        }

        if (flag) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}