HDU 4786 Fibonacci Tree

Fibonacci Tree

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

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

 

    生成树的深入理解，也就是说：

  （1）白边的最小条数（L）与最大条数（R）是一个定值 ；

  （2）黑边可以由白边替换。

 

#include <iostream>
#include <string>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
const int   Max_N = 100008  ;
struct Edge{
        int u  ;
        int v  ;
        int w  ;
} ;
Edge edge[Max_N]  ;
int N  , M;

bool cmp1(Edge A ,Edge B){
     return A.w < B.w  ;
}

bool cmp2(Edge A ,Edge B){
     return A.w > B.w  ;
}

int father[Max_N]  ;

int find_father(int  x){
      if(x == father[x])
           return  x   ;
      else
           return   father[x] = find_father(father[x])  ;
}

int gao(){
      int sum = 0  ,brige = 0 ;
      for(int i = 1 ; i <= N ; i++)
            father[i] = i ;
      for(int i = 1 ; i <= M ; i++){
            int f_u = find_father(edge[i].u)  ;
            int f_v = find_father(edge[i].v)  ;
            if(f_u != f_v){
                  brige ++  ;
                  sum += edge[i].w  ;
                  father[f_u] = f_v  ;
            }
            if(brige == N-1)
                 break  ;
      }
      return  brige == N-1 ? sum : -1 ;
}

int fibo[31]  ;

void init_fibo(){
       fibo[1] = 1 ;
       fibo[2] = 2 ;
       for(int i =3 ; i <= 30 ; i++)
              fibo[i] = fibo[i-1] + fibo[i-2]  ;
}

int judge(){
     int L  , R  ;
     sort(edge+1 ,edge+1+M, cmp1) ;
     L = gao()  ;
     sort(edge+1 ,edge+1+M ,cmp2) ;
     R = gao()  ;
     if(L == -1)
           return  0  ;
     for(int i =1 ;i < 30 ;i++){
           if(L <= fibo[i] && fibo[i] <= R)
                return  1  ;
     }
     return 0  ;
}

int main(){
    init_fibo() ;
    int T  ;
    scanf("%d",&T)  ;
    for(int cas =1 ;cas <= T; cas++){
            scanf("%d%d",&N,&M)  ;
            for(int i = 1 ;i <= M ;i++)
                  scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w)  ;
            printf("Case #%d: %s\n",cas,judge()? "Yes" : "No") ;
    }
    return 0;
}