Fibonacci Tree
Time 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?
(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).
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
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,m int 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); } } }