HDU - 4786 Fibonacci Tree (Kruskal)

　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

Sample Output

Case #1: Yes
Case #2: No

题解：用Kruskal方法求两遍生成树，一遍优先用黑边，一遍优先用白边，得到用白边的最小数量mi和最大数量ma，那么只要

[mi,ma]中有Fibonacci 数就肯定能实现。

代码：

#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;
 
const int MAXN = 100005;
 
int N,M;//点数；边数。 
 
int pre[MAXN];//记录点的祖先 
int Fi[MAXN];
 
struct Edge{//边结构体 
	int from;
	int to;
	int value;
	Edge(){}
	Edge(int a,int b,int c):from(a),to(b),value(c){}
};
 
struct cmp1{
	bool operator()(struct Edge a,struct Edge b){
		return a.value > b.value;
	}
};

struct cmp2{
	bool operator()(struct Edge a,struct Edge b){
		return a.value < b.value;
	}
};
 
priority_queue<struct Edge,vector<struct Edge>,cmp1> Q1;
priority_queue<struct Edge,vector<struct Edge>,cmp2> Q2;
 
int Find(int a){
	if(pre[a] == a)return a;
	return pre[a] = Find(pre[a]);
}
 
bool Judge(int a,int b){
	int A = Find(a);
	int B = Find(b);
	if(A != B){
		pre[A] = B;
		return true;
	}
	return false;
}
 
int Num;//表示已经找到的边数；表示最小生成树的值。 
 
inline void init(){
	while(!Q1.empty())Q1.pop(); //清空优先队列
	while(!Q2.empty())Q2.pop(); 
	Num = 0;
	for(int i=0 ; i<=N ; ++i)pre[i] = i;//初始化祖先为自己。
}
 
int main(){
	
	Fi[1] = 1;
	Fi[2] = 2;
	for(int i=3 ; Fi[i-1]<100005 ; ++i){
		Fi[i] = Fi[i-1] + Fi[i-2];
	}
	
	int T,a,b,c;
	scanf("%d",&T);
	for(int _=1 ; _<=T ; ++_){
		scanf("%d %d",&N,&M);
		init();
		for(int i=1 ; i<=M ; ++i){
			scanf("%d %d %d",&a,&b,&c);
			Q1.push(Edge(a,b,c));
			Q2.push(Edge(a,b,c));
		}
		int mi = 0,ma = 0;
		while(!Q1.empty() && Num<N-1){//找N-1条边就够了。 
			if(Judge(Q1.top().from,Q1.top().to)){
				mi += Q1.top().value;
				++Num;
			}
			Q1.pop();
		}
		if(Num < N-1){//判断是否联通 
			printf("Case #%d: No\n",_);
			continue;
		}
		Num = 0;
		for(int i=0 ; i<=N ; ++i)pre[i] = i;
		while(!Q2.empty() && Num<N-1){
			if(Judge(Q2.top().from,Q2.top().to)){
				ma += Q2.top().value;
				++Num;
			}
			Q2.pop();
		}
		printf("Case #%d: ",_);
		for(int i=1 ; i<MAXN ; ++i){
			if(Fi[i]>=mi && Fi[i]<=ma){
				printf("Yes\n");
				break;
			}
			else if(Fi[i] > ma){
				printf("No\n");
				break;
			}
		}
	}
	
	return 0;
}

HDU - 4786 Fibonacci Tree (Kruskal)

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