HDOJ 4786 Fibonacci Tree(最小生成ツリー--kruskul)
3644 ワード
Fibonacci Tree
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3004 Accepted Submission(s): 964
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
Sample Output
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3004 Accepted Submission(s): 964
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
， kruskul， ， ， fibonacci ， 100000， fibonacci ， ， fibonacci
ac ：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 1000010
#define INF 0xfffffff
#define min(a,b) (a>b?b:a)
using namespace std;
int sum,ans;
int pri[MAXN];
int fibonacci[26]={0,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025};//
struct s
{
int a;
int b;
int cost;
}dis[MAXN];
bool cmp(s A,s B)
{
return A.cost<B.cost;
}
int find(int x)
{
int r=x;
while(r!=pri[r])
r=pri[r];
int i=x,j;
while(i!=r)
{
j=pri[i];
pri[i]=r;
i=j;
}
return r;
}
void connect(int xx,int yy,int num)
{
int nx=find(xx);
int ny=find(yy);
if(nx!=ny)
{
pri[nx]=ny;
ans++;//
if(num==1)//
sum++;
}
}
int main()
{
int i,j,m,s,n;
int t,q,num,k,low,high;
int cas=0;
scanf("%d",&t);
while(t--)
{
sum=0;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)
pri[i]=i;
for(i=0;i<m;i++)
{
scanf("%d%d%d",&dis[i].a,&dis[i].b,&dis[i].cost);
}
sort(dis,dis+m,cmp);
ans=0;
for(i=0;i<m;i++)//
{
connect(dis[i].a,dis[i].b,dis[i].cost);
}
if(ans!=n-1) // ， ， No
{
printf("Case #%d: No
",++cas);
continue;
}
low=sum;
sum=0;
for(i=1;i<=n;i++)//
pri[i]=i;
for(i=m-1;i>=0;i--)//
{
connect(dis[i].a,dis[i].b,dis[i].cost);
}
high=sum;
int bz=0;
for(i=1;i<=24;i++)
{
if(fibonacci[i]>=low&&fibonacci[i]<=high)
bz=1;
}
if(bz)
printf("Case #%d: Yes
",++cas);
else
printf("Case #%d: No
",++cas);
}
return 0;
}