hdu4786（最小生成树）

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3820 Accepted Submission(s): 1214



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 <= 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

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

题意：有一个图只存在白边和黑边，现在要求一个生成树，问生成树的白边的数量是否可以是一个斐波拉契数。

思路：将所有白边优先建最小生成树得到白边的最多数，在将所有黑边优先建最小生成树，得到白边的最少数，然后在这个区间看是否存在斐波那契数。（图要联通，不连通直接No）

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
const int N=100010;
typedef long long ll;
int m,n;
struct data
{
int a,b,w;
} tu
;

int f
;

bool cmp1(data a,data b)
{
return a.w>b.w;
}

bool cmp2(data a,data b)
{
return a.w<b.w;
}

int fa(int x)
{
if(x!=f[x])
return f[x]=fa(f[x]);
return f[x];
}

int main()
{
int t,o=1;
cin>>t;
while(t--)
{
scanf("%d%d",&n,&m);
for(int i=0; i<m; i++)
scanf("%d%d%d",&tu[i].a,&tu[i].b,&tu[i].w);
sort(tu,tu+m,cmp1);
for(int i=1; i<=n; i++)
f[i]=i;
int l=0;
for(int i=0; i<m; i++)
{
int f1=fa(tu[i].a),f2=fa(tu[i].b);
if(f1!=f2)
{
f[f1]=f2;
if(tu[i].w==1)l++;
}
}
sort(tu,tu+m,cmp2);
for(int i=1; i<=n; i++)
f[i]=i;
int r=0;
for(int i=0; i<m; i++)
{
int f1=fa(tu[i].a),f2=fa(tu[i].b);
if(f1!=f2)
{
f[f1]=f2;
if(tu[i].w==1)r++;
}
}
if(r<l)swap(l,r);
int ff=fa(1),flag=1;
for(int i=2; i<=n; i++)
if(fa(i)!=ff)
{
flag=0;
break;
}
if(!flag)
printf("Case #%d: No\n",o++);
else
{
int f0=0,f1=1,f2;
for(int i=0;; i++)
{
f2=f1+f0;
if(f2>=l&&f2<=r)
{
printf("Case #%d: Yes\n",o++);
break;
}
else if(f2>r)
{
printf("Case #%d: No\n",o++);
break;
}
f0=f1,f1=f2;
}
}
}
return 0;
}