hdu 4786 Fibonacci Tree

hdu 4786 Fibonacci Tree"/>

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;elsecout << "No" << endl;}return 0;
}