HDOJ question 4786 maid Tree (cruise Karl)

HDOJ question 4786 maid Tree (cruise Karl)
Fibonacci TreeTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 2562 Accepted Submission (s): 816


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?
(Maid 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

24 41 2 12 3 13 4 11 4 05 61 2 11 3 11 4 11 5 13 5 14 2 1

Sample Output

Case #1: YesCase #2: No

Source 2013 Asia Chengdu Regional Contest
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#include
 
  #include
  
   #include
   
    struct s{int u,v,w;}edge[100100];int pre[100100],a[100100],n,m;int cmp1(const void *a,const void *b){return (*(struct s *)a).w-(*(struct s *)b).w;}int cmp2(const void *a,const void *b){return (*(struct s *)b).w-(*(struct s *)a).w;}void init(int n){int i;for(i=0;i<=n;i++)pre[i]=i;}void fun(){int i;a[1]=1;a[2]=2;for(i=3;i<100100;i++)a[i]=a[i-1]+a[i-2];}int find(int x){if(x==pre[x])return pre[x];return pre[x]=find(pre[x]);}int ku(){init(n);int ans=0,i,j;for(i=0;i
    
     1)return -1;return ans;}int main(){int t,c=0;scanf("%d",&t);fun();while(t--){//int n,mint i,l,r;scanf("%d%d",&n,&m);//init(n);for(i=0;i
     
      r)break;if(a[i]<=r&&a[i]>=l){printf("Case #%d: Yes\n",++c);flag=1;break;}}if(!flag)printf("Case #%d: No\n",++c);}}}