Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3006    Accepted Submission(s): 966

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

题解：

这个题就是求一遍最小的生成树，

求一遍最大的生成树

两个中间是不是有斐波那契数

最后还有判一下联通；

kruskal;

代码：

  1 #include<stdio.h>
  2 #include<stdlib.h>
  3 #include<string.h>
  4 #include<algorithm>
  5 using namespace std;
  6 const int MAXN=100010;
  7 struct Node {
  8     int s,e,c;
  9 };
 10 Node dt[MAXN];
 11 int pre[MAXN];
 12 int M,t1,N;
 13 int cmp1(Node a,Node b){
 14     return a.c<b.c;
 15 }
 16 int cmp2(Node a,Node b){
 17     return a.c>b.c;
 18 }
 19 /*int cmp1(const void *a,const void *b){
 20     if((*(Node *)a).c<(*(Node *)b).c)return -1;
 21     else return 1;
 22 }
 23 int cmp2(const void *a,const void *b){
 24     if((*(Node *)a).c>(*(Node *)b).c)return -1;
 25     else return 1;
 26 }*/
 27 int find(int x){
 28     return pre[x]= x==pre[x]?x:find(pre[x]);
 29 }
 30 bool merge(Node a){
 31     if(!pre[a.s])pre[a.s]=a.s;
 32     if(!pre[a.e])pre[a.e]=a.e;
 33     int f1,f2;
 34     f1=find(a.s);f2=find(a.e);
 35     if(f1!=f2){
 36         pre[f1]=f2;
 37         t1++;
 38         if(a.c)return true;
 39     }
 40     return false;
 41 }
 42 int kruskal(){int tot=0;
 43         t1=1;
 44     for(int i=0;i<M;i++){
 45         if(merge(dt[i]))tot++;
 46     }
 47     if(t1==N)return tot;
 48     else return -1;
 49 }
 50 bool fp[MAXN];
 51 void gf(){
 52     int a,b,c=0;
 53     memset(fp,false,sizeof(fp));
 54     a=1;b=2;
 55     fp[a]=fp[b]=true;
 56     while(c<MAXN){
 57         c=a+b;
 58         fp[c]=true;
 59         a=b;
 60         b=c;
 61     }
 62 }
 63 int main(){
 64     int T,s1,s2,ans,flot=0;
 65     scanf("%d",&T);
 66     while(T--){
 67             flot++;
 68             memset(pre,0,sizeof(pre));
 69         scanf("%d%d",&N,&M);
 70         for(int i=0;i<M;i++){
 71             scanf("%d%d%d",&dt[i].s,&dt[i].e,&dt[i].c);
 72         }
 73        // qsort(dt,M,sizeof(dt[0]),cmp1);
 74        sort(dt,dt+M,cmp1);
 75         s1=kruskal();
 76         //qsort(dt,M,sizeof(dt[0]),cmp2);
 77         sort(dt,dt+M,cmp2);
 78         memset(pre,0,sizeof(pre));
 79         s2=kruskal();
 80         //printf("%d %d\n",s1,s2);
 81         gf();
 82         ans=0;
 83         if(s1<0||s2<0){
 84             printf("Case #%d: No\n",flot);
 85             continue;
 86         }
 87        //for(int i=0;i<100;i++)printf("fp[%d]=%d ",i,fp[i]);puts("");
 88         if(s1>s2){
 89             int q=s1;
 90             s1=s2;
 91             s2=q;
 92         }
 93         for(int i=s1;i<=s2;i++){
 94             if(fp[i])ans=1;
 95         }
 96         if(ans)printf("Case #%d: Yes\n",flot);
 97         else printf("Case #%d: No\n",flot);
 98     }
 99     return 0;
100 }