# Fibonacci Tree

## 题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=4786

## 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 <= 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.

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

Case #1: Yes
Case #2: No

## 代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
struct node{
int x,y,z;
}p[maxn];
bool cmp1(node A,node B){
return A.z<B.z;
}
bool cmp2(node A,node B){
return A.z>B.z;
}
int fa[maxn];
int fi(int x){
return x==fa[x]?x:fa[x]=fi(fa[x]);
}
void uni(int x,int y){
x=fi(x),y=fi(y);
if(x==y)return;
else fa[x]=fa[y];
}
void solve(int Cas){
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
fa[i]=i;
for(int i=1;i<=m;i++)
scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);
sort(p+1,p+1+m,cmp1);
int x1=0,x2=0;
int D = 0;
for(int i=1;i<=m;i++){
if(fi(p[i].x)!=fi(p[i].y)){
x1+=p[i].z;
uni(p[i].x,p[i].y);
D = D + 1;
}
}
if(D!=n-1){
printf("Case #%d: No\n",Cas);
return;
}
for(int i=1;i<=n;i++)
fa[i]=i;
sort(p+1,p+1+m,cmp2);
for(int i=1;i<=m;i++){
if(fi(p[i].x)!=fi(p[i].y)){
x2+=p[i].z;
uni(p[i].x,p[i].y);
}
}
long long a=1,b=1;
int flag = 0;
while(a<=x2||b<=x2){
if(a>=x1&&a<=x2)
flag=1;
if(b>=x1&&b<=x2)
flag=1;
a=a+b;
if(a>b)swap(a,b);
}
if(a>=x1&&a<=x2)
flag=1;
if(b>=x1&&b<=x2)
flag=1;
if(flag)printf("Case #%d: Yes\n",Cas);
else printf("Case #%d: No\n",Cas);
}
int main(){
int t;
scanf("%d",&t);
for(int cas=1;cas<=t;cas++)solve(cas);
}
posted @ 2017-07-11 16:15 qscqesze 阅读(...) 评论(...) 编辑 收藏