HDU 4786 Fibonacci Tree（最小生成树变式）

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

 

题目大意：给定一个图，里面有的边是白色，剩下的全为黑色。问存不存在一个生成树里面包含的白色边的个数恰为Fibonacci numbers。

看了题解始终没明白。。。题解的解法是白边边权为 1 ，黑边边权为 0 。构造一个最大生成树，用掉的白边数就是所有生成树里白边的最大个数Max。再构造一个最小生树，用掉的白边数就是所有生成树里白边的最小个数Min。若Min~Max之间包含 Fibonacci numbers 就符合题意。据说是因为一条黑边可以被一条白边代替，依然保持图的连通性，所以只要Min ~ Max之间包含 Fibonacci numbers 就OK。依然不明白是为什么。。。求大神告知。生成树用 Kruskal，查找Fibonacci numbers 是打了一个表然后二分找的。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1e5 + 5;int p[maxn],f[100],v[maxn],u[maxn],e[maxn],w[maxn];int len = 1;int n,m;bool cmp(int x,int y){    return w[x] < w[y];}void init(){    memset(f,0,sizeof(f));    f[0] = 1,f[1] = 2;    while(f[len] <= 100000) ++len,f[len] = f[len - 1] + f[len - 2];}int Find(int x){    return p[x] == x ? x : p[x] = Find(p[x]);}int main(){    int t,x,y,c;    scanf("%d",&t);    init();    int pre = t;    while(t--)    {        scanf("%d %d",&n,&m);        for(int i = 1;i <= m; ++i) e[i] = i;        for(int i = 1;i <= m; ++i) scanf("%d %d %d",&u[i],&v[i],&w[i]);        sort(e + 1,e + m + 1,cmp);        int temp = 0,mi = 0,ma = 0;        for(int i = 1;i <= n; ++i) p[i] = i;        for(int i = 1;i <= m; ++i)        {            int now = e[i];            int x = Find(u[now]),y = Find(v[now]);            if(x != y)            {                ++temp;                if(w[now] == 1) ++mi;                p[y] = x;            }        }        if(temp <  n - 1)        {            printf("Case #%d: No\n",pre - t);            continue;        }        for(int i = 1;i <= n; ++i) p[i] = i;        for(int i = m;i >= 1; --i)        {            int now = e[i];            int x = Find(u[now]),y = Find(v[now]);            if(x != y)            {                if(w[now] == 1) ++ma;                p[y] = x;            }        }        temp = lower_bound(f,f + len,mi) - f;        printf("Case #%d: ",pre - t);        if(f[temp] <= ma) printf("Yes\n");        else printf("No\n");    }    return 0;}