HDU - 4786 - Fibonacci Tree

---

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4786

题目描述

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, … )

一个图，N个顶点，M条边，每条边或为白色或为黑色。 问你能否用两色边构造出一颗生成树，使树中白色边的数量为一个Fibonacci数。 ( Fibonacci数定义为 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). 第一行输入一个整数T，表示样例个数。

每个样例第一行包括两个整数 N(1 <= N <= 10^5) and M(0 <= M <= 10^5) ，分别表示点和边的数量。 接下来 M 行，每行包括三个整数 u, v (1 <= u,v <= N, u<> v)和 c (0 <= c <= 1)，表示u和v之间有一条颜色为c的边（1表示白色，0表示黑色）。

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.

对于每个样例，输出一行“Case #x: s”。x表示第几个样例， s表示答案（“Yes” 或 “No”）。

Sample Input

2 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 Output

Case #1: Yes Case #2: No

解题思路

先说一下题意，连接所有的点，最后生成 生成树 ，连接的时候用的边不是白的就是黑的（人为规定），问用的白边的条数是否是一个Fibonacci数 第一点，没说是最小生成树 第二点，存在就输出yes 明白了这个点，那么我们把最小生成树用的白边和最大生成树的白边，都算出来，如果这里面有Fibonacci数的话，那么就yes，所以后面控制黑白的数字还是有用的，到时候直接加上就ok

AC代码

#include<iostream> #include<algorithm> #include<cstring> #define maxn 100010 using namespace std; int n, m; int pre[maxn]; int use[40]; struct node { int a, b, c; bool operator < (const node& rhs) const { return c < rhs.c; } }cnt[maxn]; int find(int x) { return pre[x] == x ? x : pre[x] = find(pre[x]); } bool kruskal() { int num = 1, ans1 = 0, ans2 = 0; for(int i = 1; i <= n; i++) pre[i] = i; sort(cnt, cnt+m); for(int i = 0; i < m; i++) { int x = find(cnt[i].a); int y = find(cnt[i].b); if(x != y) { pre[y] = x; num++; ans1 += cnt[i].c; } } if(num != n) return false; for(int i = 1; i <= n; i++) pre[i] = i; for(int i = m-1; i >= 0; i--) { int x = find(cnt[i].a); int y = find(cnt[i].b); if(x != y) { pre[y] = x; ans2 += cnt[i].c; } } for(int i = 1; i <= 26; i++) if(ans1 <= use[i] && use[i] <= ans2) return true; return false; } void init() { use[1] = 1, use[2] = 2; for(int i = 3; i <= 26; i++) use[i] = use[i-1] + use[i-2]; } int main() { init(); int t, kase = 1; scanf("%d", &t); while(t--) { scanf("%d %d" , &n, &m); for(int i = 0; i < m; i++) scanf("%d %d %d", &cnt[i].a, &cnt[i].b, &cnt[i].c); if(kruskal()) printf("Case #%d: Yes\n", kase++); else printf("Case #%d: No\n", kase++); } return 0; }