HDU 4786 Fibonacci Tree（最小生成树变式）

最新推荐文章于 2021-07-28 14:29:18 发布

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最新推荐文章于 2021-07-28 14:29:18 发布

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分类专栏：图论算法

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探讨如何通过构建最大与最小生成树确定是否能形成含有Fibonacci数个白边的生成树。使用Kruskal算法并结合二分查找验证。

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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

Sample Output

Case #1: Yes
Case #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;
}