HDU 4786 Fibonacci Tree （2013成都1006题）

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 75    Accepted Submission(s): 38

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

 

Source

2013 Asia Chengdu Regional Contest

 

只要白边优先和黑边优先两种顺序做两次最小生成树。

得到白边数量的区间，然后枚举斐波那契数列就可以了。

注意如果一开始是非连通的，输出NO

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013-11-16 14:14:50
 File Name     :E:\2013ACM\专题强化训练\区域赛\2013成都\1006.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 int f[];

 struct Edge
 {
     int u,v,c;
 }edge[];
 int F[];
 int find(int x)
 {
     if(F[x] == -)return x;
     return F[x] = find(F[x]);
 }

 bool cmp1(Edge a,Edge b)
 {
     return a.c < b.c;
 }
 bool cmp2(Edge a,Edge b)
 {
     return a.c > b.c;
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int tot = ;
     f[] = ;
     f[] = ;
     while(f[tot] <= )
     {
         f[tot+] = f[tot] + f[tot-];
         tot++;
     }
     int T;
     int iCase = ;
     int n,m;
     scanf("%d",&T);
     while(T--)
     {
         iCase++;
         scanf("%d%d",&n,&m);
         for(int i = ;i < m;i++)
             scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].c);
         sort(edge,edge+m,cmp1);
         memset(F,-,sizeof(F));
         int cnt = ;
         for(int i = ;i < m;i++)
         {
             int t1 = find(edge[i].u);
             int t2 = find(edge[i].v);
             if(t1 != t2)
             {
                 F[t1] = t2;
                 if(edge[i].c == )cnt++;
             }
         }
         int Low = cnt;
         memset(F,-,sizeof(F));
         sort(edge,edge+m,cmp2);
         cnt = ;
         for(int i = ;i < m;i++)
         {
             int t1 = find(edge[i].u);
             int t2 = find(edge[i].v);
             if(t1 != t2)
             {
                 F[t1] = t2;
                 if(edge[i].c == )cnt++;
             }
         }
         int High = cnt;
         bool ff = true;
         for(int i = ;i <= n;i++)
             if(find(i) != find())
             {
                 ff = false;
                 break;
             }
         if(!ff)
         {
             printf("Case #%d: No\n",iCase);
             continue;
         }
         bool flag = false;
         for(int i = ;i <= tot;i++)
             if(f[i] >= Low && f[i] <= High)
                 flag = true;
         if(flag)
             printf("Case #%d: Yes\n",iCase);
         else printf("Case #%d: No\n",iCase);

     }
     return ;
 }