Fibonacci tree (minimum spanning trees, maximum spanning tree)

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 when Uncle Yang wants him to doing some on Spann ing Tree. So Coach Pang decides to solve the following problem:
Consider a bidirectional graph G with N vertices and M edges. All edges is painted into the 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 the integers n (1 <= n <=) 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), Indic Ating an edge between U and V with a color C (1 for white and 0 for black).

Output a line ' case #x: s ' for each test case. X is the case number and s are either "Yes" or "No" (without quotes) representing the answer to the problem.

Sample Input2 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 outputcase #1: Yes case #2: The problem is to ask for the smallest spanning tree, to find the largest spanning tree in the middle of two is not the Fibonacci number and finally sentenced to unicom; Kruskal; code:

1#include <stdio.h>2#include <stdlib.h>3#include <string.h>4#include <algorithm>5 using namespacestd;6 Const intmaxn=100010;7 structNode {8     ints,e,c;9 };Ten Node DT[MAXN]; One intPRE[MAXN]; A intM,t1,n; - intCmp1 (Node a,node b) { -     returna.c<B.C; the } - intCMP2 (Node a,node b) { -     returnA.c>B.C; - } + /*int cmp1 (const void *a,const void *b) { - if ((* (node *) a) .c< (* (node *) b). c) return-1; + else return 1; A } at int cmp2 (const void *a,const void *b) { - if ((* (node *) a) .c> (* (node *) b). c) return-1; - else return 1; - }*/ - intFindintx) { -     returnpre[x]= x==pre[x]?X:find (pre[x]); in } - BOOLmerge (Node a) { to     if(!pre[a.s]) pre[a.s]=A.S; +     if(!PRE[A.E]) pre[a.e]=A.E; -     intF1,f2; theF1=find (A.S); f2=find (A.E); *     if(f1!=F2) { $pre[f1]=F2;Panax Notoginsengt1++; -         if(A.C)return true; the     } +     return false; A } the intKruskal () {inttot=0; +t1=1; -      for(intI=0; i<m;i++){ $         if(merge (Dt[i])) tot++; $     } -     if(t1==n)returntot; -     Else return-1; the } - BOOLFP[MAXN];Wuyi voidGF () { the     intA,b,c=0; -memset (FP,false,sizeof(FP)); WuA=1; b=2; -fp[a]=fp[b]=true; About      while(c<MAXN) { $c=a+b; -fp[c]=true; -A=b; -b=C; A     } + } the intMain () { -     intt,s1,s2,ans,flot=0; $scanf"%d",&T); the      while(t--){ theflot++; thememset (PRE,0,sizeof(pre)); thescanf"%d%d",&n,&M); -          for(intI=0; i<m;i++){ inscanf"%d%d%d",&dt[i].s,&dt[i].e,&dt[i].c); the         } the        //qsort (dt,m,sizeof (dt[0]), CMP1); AboutSort (dt,dt+m,cmp1); thes1=Kruskal (); the         //qsort (dt,m,sizeof (dt[0]), CMP2); theSort (dt,dt+m,cmp2); +memset (PRE,0,sizeof(pre)); -S2=Kruskal (); the         //printf ("%d%d\n", s1,s2);Bayi GF (); theans=0; the         if(s1<0|| s2<0){ -printf"Case #%d:no\n", flot); -             Continue; the         } the        //for (int i=0;i<100;i++) printf ("fp[%d]=%d", I,fp[i]);p UTS (""); the         if(s1>S2) { the             intq=S1; -s1=S2; theS2=Q; the         } the          for(inti=s1;i<=s2;i++){94             if(Fp[i]) ans=1; the         } the         if(ANS) printf ("Case #%d:yes\n", flot); the         Elseprintf"Case #%d:no\n", flot);98     } About     return 0; -}

Fibonacci tree (minimum spanning trees, maximum spanning tree)