hdoj 4786 Fibonacci Tree

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Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3583 Accepted Submission(s): 1152

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

24 41 2 12 3 13 4 11 4 05 61 2 11 3 11 4 11 5 13 5 14 2 1

Sample Output

Case #1: YesCase #2: No
题意：
    题目要求  1 表示白线连接，0 表示黑线连接！
             图要连通，但所构成图的白色线个数必须符合是属于fib序列！并不需要所有的线全都是白线！
    思路：    
             可以根据线条的颜色，先尽量用白色线条进行图的连通，求出满足连通时白色线条的条数max ，相反，尽量用黑色线条，求出
             此时满足连通的白色线条的个数min，再min和max之间寻找是否有fib满足数列的数！
     具体看代码：
          #include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define  N  110000int set[N],n,m;int fib[1000000+2100];struct node{int start;int end;int judge;}num[N];bool cmp1(node a,node  b)//按照先黑再白顺序 {return  a.judge <b.judge;}bool cmp2(node a,node  b)//顺序与前面相反 {return  a.judge>b.judge;}void  dabiao(){  fib[1]=1;fib[2]=2;for(int  i=3;i<1000000+100;i++){fib[i]=fib[i-1]+fib[i-2];}}int find(int  p)//找根节点 {int t;int child=p;while(p!=set[p])p=set[p];while(child!=p){t=set[child];set[child]=p;child=t;}return  p;}void merge(int x,int y)//加入集合 {int  fx=find(x);int fy=find(y);if(fx!=fy)set[fx]=fy;}int main(){int  t,k=1;memset(fib,0,sizeof(fib));dabiao();scanf("%d",&t);while(t--){int i,j,n,m,max=0,min=0;scanf("%d%d",&n,&m);for(i=1;i<=m;i++)scanf("%d%d%d",&num[i].start ,&num[i].end ,&num[i].judge );for(i=1;i<=n;i++)    set[i]=i;sort(num+1,num+m,cmp2); //尽量先用白色线条！ for(i=1;i<=m;i++){if(find(num[i].start )!=find(num[i].end )){merge(num[i].start ,num[i].end );if(num[i].judge ) max++;}} for(i=1;i<=n;i++)    set[i]=i;sort(num+1,num+m,cmp1);//尽量先用黑色线条！ for(i=1;i<=m;i++){if(find(num[i].start )!=find(num[i].end)){merge(num[i].start ,num[i].end );if(num[i].judge )min++;}} printf("Case #%d: ",k++);int exit=0;for(i=1;i<=n;i++)//看是否连通！ {if(set[i]==i) exit++; if(exit>1) break;}if(exit>1){printf("No\n");continue;}exit=0;for(i=1;fib[i]<=max;i++){if(fib[i]>=min&&fib[i]<=max){exit++;}if(exit)break;}if(exit)printf("Yes\n");elseprintf("No\n");}return  0;}

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