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 <= 10 5) and M(0 <= M <= 10 5).
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 1Sample Output
Case #1: Yes
Case #2: No题意：给你n个点，m个关系，1 表示白边， 0表示黑边，为是否有 k条白边和随便几条黑边构成一颗树，并且k要为斐波那契数；
思路就是黑边不限定数量 但是限制的是白边的数量一定要是斐波那契数 ，并且只有1， 0两种边；
那是不是判断一个最多的白边构成的树，和一个最少白边构成的树，在这期间的白边数一定能构成树；
你想最大生成树也就是最多白边构成的树，那如果我现在减少一条边，是不是后有一个点空出来，那我现在重新构成树，但是现在我不把拿出来的白边算进去，是不是后构成新的树 而那个点会有别的边相连； 树可能不是同一棵树，但是题目树一定能构成树就对了；
你这是要特判一下 本来就不是树的情况；
ac码
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
using namespace std;
typedef long long LL;
int fa[100005];
struct po
{
int u ,v ,w;
}eg[100005];
int n,m;
int fb() //预处理斐波那契数列
{
a[0] = 1,a[1] = 2;
for(int i = 2; ; i++)
{
a[i] = a[i - 1] + a[i - 2];
if(a[i] > 100500)
{
return i;
break;
}
}
}
bool cmp1(po x,po y)
{
return x.w < y.w; //xiao到da
}
bool cmp2(po x,po y)
{
return x.w > y.w; //da到xiao
}
int find(int x)
{
if(x!=fa[x])
fa[x]=find(fa[x]);
return fa[x];
}
int bpf() //避圈法
{
int tot = n;
int sum = 0;
for(int i = 0;i < m;i++)
{
int x = find(eg[i].u);
int y = find(eg[i].v);
if(x == y)
continue;
fa[x] = y;
sum += eg[i].w;
tot--; //因为会有重边 其实我感觉不需要也行
if(tot == 0)break;
}
return sum;
}
int main()
{
int t;
int fbb = fb();
scanf("%d",&t);
for(int k = 1;k <= t;k++)
{
scanf("%d %d",&n,&m);
for(int i = 1;i <= m;i++)
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
eg[i].u = a;
eg[i].v = b;
eg[i].w = c;
}
for(int i = 1;i <= n;i++) fa[i] = i;
sort(eg + 1,eg + m + 1,cmp2);
int r = bpf(); // 最大白边生成树
for(int i = 1;i <= n;i++) fa[i] = i;
sort(eg + 1, eg + m + 1 , cmp1);
int l = bpf(); //最小白边生成树
int ff = 1;
for(int i = 1;i <= n;i++) //特判一下是不是n个点一定能构成一棵树
if(find(i) != find(1))
{
ff = 0;
break;
}
if(ff == 0) //不能构成一棵树
{
printf("Case #%d: No\n",k);
continue;
}
int flag = 0;
for(int i = 1;i < fbb;i++) //判断期间是否存在斐波那契数
if(a[i] >= l && a[i] <= r)
flag = 1;
if(flag)
printf("Case #%d: Yes\n",k);
else
printf("Case #%d: No\n",k);
}
return 0;
}