题意：

给定一个无向图

n 个点 m条无向边

u v val

val == 1 表示边(u, v) 为白边

问能否找到n个点的生成树， 使得白边数为斐波那契数

思路：

并查集求图是否连通（ 是否存在生成树）

求出 最多白边树 的 白边数量

求出 最少白边树 的 白边数量

若[最少， 最多] 区间内存在斐波那契数 ，则满足条件

（也就是说，白边的数量是连续有解的）

//YY得证

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<queue>
#include<string.h>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<math.h>
#include<algorithm>
#define N 101010
#define inf 10000000
using namespace std;
inline int Min(int a,int b){return a>b?b:a;}
inline int Max(int a,int b){return a<b?b:a;}int f[N];
int find(int x){return x==f[x]?x:(f[x] = find(f[x]));}
void Union(int u, int v){int fu = find(u), fv = find(v);if(fu>fv)f[fu] = fv;else f[fv] = fu;
}
set<int>fib;
int n, m;
struct node{int u,v,c;
}edge[N];
int edgenum;
bool cmp1(node a,node b){return a.c<b.c;}
bool cmp2(node a,node b){return a.c>b.c;}
int main(){int T, Cas = 1;scanf("%d",&T);int i, j, col;fib.clear();fib.insert(1);fib.insert(2);j=1;for(i=2;i<=N;){fib.insert(i+j);int lala = i;i = i+j;j = lala;}while(T--){scanf("%d %d", &n, &m);for(i=1;i<=n;i++)f[i] = i;edgenum = 0;while(m--){int u,v;scanf("%d %d %d",&u,&v,&col);edge[edgenum].u = u;edge[edgenum].v = v;edge[edgenum++].c = col;int fx = find(u), fy = find(v);if(fx == fy)continue;Union(fx,fy);}printf("Case #%d: ",Cas++);for(i=1;i<=n;i++)find(i);bool su = true;for(i=1;i<=n;i++)if(f[i]!=f[1]){su = false; break;}if(su == false){printf("No\n");continue;}for(i=1;i<=n;i++)f[i] = i;sort(edge, edge+edgenum, cmp1);int size = 0, bl=0, bm=0;for(i=0;i<edgenum;i++){int u =edge[i].u, v=edge[i].v;int fu = find(u), fv=find(v);if(fu == fv)continue;size++;bl+= edge[i].c;Union(fu, fv);if(size==n-1)break;}for(i=1;i<=n;i++)f[i] = i;sort(edge, edge+edgenum, cmp2);size = 0;for(i=0;i<edgenum;i++){int u =edge[i].u, v=edge[i].v;int fu = find(u), fv=find(v);if(fu == fv)continue;size++;bm+= edge[i].c;Union(fu, fv);if(size==n-1)break;}if(fib.upper_bound(bl) == fib.end() ){printf("No\n");continue;}if((*fib.lower_bound(bl) )>bm){printf("No\n");continue;}printf("Yes\n");}return 0;
}
/*
4 2
1 2 1
3 4 1*/