hdu 4786 Fibonacci Tree

题目链接：hdu4786

题意：给你一个图，判断是否存在一个生成树，使得权值之和为一个斐波那契数。

思路：分别跑一次最大生成树和最小生成树。然后判断两个生成树的权值中间是否存在一个斐波那契数。如果存在则输出yes，否则输出no

代码：

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;

struct node {
	int from, to, w;
	node(int a, int b, int c) :from(a), to(b), w(c) {}
};

int vex[maxn];
int n, m;

int fib[maxn];
vector<node> v;

void fibs() {
	int i;
	fib[0] = 1;
	fib[1] = 1;
	for (i = 2; i < 100000; i++)
		fib[i] = fib[i - 1] + fib[i - 2];
}


void init() {
	int i;
	for (i = 0; i <= n; i++) {
		vex[i] = i;
	}
}

int find(int x)
{
	if (vex[x] == x)
		return x;
	return vex[x] = find(vex[x]);
}


void kruskal(int &cnt) {
	int i;
	int a, b;
	for (i = 0; i < m; i++) {
		auto e = v[i];
		a = e.from;
		b = e.to;
		int r1 = find(a);
		int r2 = find(b);
		if (r1 != r2) {
			vex[r1] = r2;
			cnt += e.w;
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	fibs();
	int t;
	cin >> t;
	int cases = 0;
	while (t--) {
		v.clear();
		cases++;
		cin >> n >> m;
		int a, b, w, i;
		for (i = 0; i < m; i++) {
			cin >> a >> b >> w;
			v.push_back(node(a, b, w));
		}
		sort(v.begin(), v.end(), [=](node n1, node n2)
			{
				return n1.w > n2.w;
			});
		memset(vex, 0, sizeof(vex));
		int flag = 0;
		int maxs, mins;
		maxs = mins = 0;
		init();
		kruskal(maxs);
		memset(vex, 0, sizeof(vex));
		sort(v.begin(), v.end(), [=](node n1, node n2)
			{
				return n1.w < n2.w;
			});
		init();
		kruskal(mins);
		int cnts = 0;

		for (i = 0; maxs >= fib[i]; i++) {
			if (mins <= fib[i] && maxs >= fib[i]) {
				flag = 1;
				break;
			}
		}
		for (i = 1; i <= n; i++) {
			if (vex[i] == i)
				cnts++;
			if (cnts > 1)
				break;
		}

		if (cnts > 1)
			flag = 0;
		cout << "Case #" << cases << ": ";
		if (flag)
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}
	return 0;
}