最大生成树夹最小生成树，老题目了，依稀记得当年在成都靠这题捡了个铜。。。。。
Fibonacci Tree
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1572 Accepted Submission(s): 479
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).
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 1
Sample Output
Case #1: Yes
Case #2: No
Source 2013 Asia Chengdu Regional Contest
#include
#include
#include
#include
using namespace std; const int maxn=100100; int nf,fib[100]; int getFib() { fib[0]=1;fib[1]=2; nf=2; for(int i=2;fib[nf-1]<=100100;i++) { fib[nf]=fib[nf-1]+fib[nf-2]; nf++; } } int n,m; int fa[maxn]; int find(int x) { if(x==fa[x]) return x; return fa[x]=find(fa[x]); } struct Edge { int u,v,c; }edge[maxn]; bool cmp1(Edge x,Edge y) { return x.c
y.c; } int Kruscal() { int cnt=n,ans=0; for(int i=0;i<=n+1;i++) fa[i]=i; for(int i=0;i
=MiMST&&fib[i]<=MxMST) { flag=true; break; } } printf("Case #%d: ",cas++); if(flag) puts("Yes"); else puts("No"); } return 0; }