Fibonacci Tree最小最大生成树

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7969    Accepted Submission(s): 2409

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

 

解题思路: 白边最小生成树值low  最大生成树high   中间的值生成树必定存在;

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 int t,n,m,cnt;
 8 const int N=1e5+5;
 9 int arr[N];
10 int brr[N];
11 
12 struct Node{
13     int u,v,ee;
14 }A[N];
15 
16 int judge(){
17     brr[1]=1,brr[2]=2;
18     for(int i=3;i<=31;i++){
19         brr[i]=brr[i-1]+brr[i-2];
20     }
21 }
22 
23 int cmp1(Node a,Node b){ return a.ee<b.ee; }
24 int cmp2(Node a,Node b){ return a.ee>b.ee; }
25 
26 int find_root(int x){ return arr[x]==x?x:arr[x]=find_root(arr[x]); }
27 void merge_unit(int x,int y){
28     int xx=find_root(x);
29     int yy=find_root(y);
30     if(xx!=yy) arr[xx]=yy;
31 }
32 int kurscul(){
33     cnt=0;
34     int num=0;
35     for(int i=1;i<=N-1;i++) arr[i]=i;
36     for(int i=1;i<=m;i++){
37         if(find_root(A[i].u)!=find_root(A[i].v)){
38             cnt++;
39             merge_unit(A[i].u,A[i].v);
40             num+=A[i].ee;
41         }
42     }
43     return num;
44 }
45 
46 int main(){
47     ios::sync_with_stdio(false);
48     judge();
49     int jishu=0;
50     cin>>t;
51     while(t--){
52         cin>>n>>m;
53         for(int i=1,d1,d2,d3;i<=m;i++){
54             cin>>d1>>d2>>d3;
55             A[i]={d1,d3};
56         }
57         sort(A+1,A+1+m,cmp1);
58         int low=kurscul();
59         cout << "Case #" << ++jishu << ": ";
60         if(cnt!=n-1) {cout << "No" << endl;continue;}
61         sort(A+1,cmp2);
62         int high=kurscul();
63         int rr=1;
64         while(brr[rr]<low) rr++;   //
65         if(brr[rr]<=high) cout << "Yes" << endl;
66         else cout << "No" << endl;
67     }
68     return 0;
69 }

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