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2024-08-03 00:05:32   217人阅读

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2340    Accepted Submission(s): 748


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
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
 

Source
2013 Asia Chengdu Regional Contest





题意：问构成的生成树其中是否存在黑色边（边为1）数为斐波那契数
思路：求出生成树中最小包含的黑色边数，和最多黑色边数，如果有斐波那契数在两者之间，则可以构成，因为黑白边可以搭配使用
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int n,m;
int fibo[50];
int f[100010];
struct node
{
    int u,v,c;
} s[100010];

bool cmp1(node x , node y)
{
    return x.c < y.c;
}

bool cmp2(node x, node y)
{
    return  x.c > y.c;
}

int find(int x)
{
    return x == f[x] ? x : f[x] = find(f[x]);
}

void Union(int x ,int y)
{
    int fx = find(x);
    int fy = find(y);

    if(fx != fy)
    {
        f[fx] = fy;
    }
}

int main()
{
#ifdef xxz
    freopen("in.txt","r",stdin);
#endif
    fibo[1] = 1;
    fibo[2] = 2;
    for(int i = 3; ; i++)
    {
        fibo[i] = fibo[i-1] + fibo[i-2];
        if(fibo[i] >= 100000) break;
    }

    int T,Case = 1;;
    scanf("%d",&T);

    while(T--)
    {


        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++) f[i] = i;

        for(int i = 0; i < m; i++)
        {

            scanf("%d%d%d",&s[i].u,&s[i].v,&s[i].c);
            Union(s[i].u,s[i].v);
        }
        int cent = 0;
        int bl = 0, bh = 0;
        int root = 0,size = 0;

        for(int i = 1; i <= n; i++)
        {
            if(f[i] == i)
            {
                cent++;
                root = i;
            }
        }

        printf("Case #%d: ",Case++);
        if(cent >= 2) cout<<"No"<<endl;//首先要判断是否能构成一个生成树，判断根节点个数是否为1就行
        else
        {
            sort(s,s+m,cmp1);
            for(int i = 1; i <= n; i++) f[i] = i;

            for(int i = 0; i < m; i++)
            {
                int fu = find(s[i].u);
                int fv = find(s[i].v);
                if(fu == fv) continue;

                bl += s[i].c;
                size++;
                Union(s[i].u,s[i].v);
                if(size == n-1) break;
            }

            size = 0;
            sort(s,s+m,cmp2);
            for(int i = 1; i <= n; i++) f[i] = i;


            for(int i = 0; i < m; i++)
            {
                int fu = find(s[i].u);
                int fv = find(s[i].v);
                if(fu == fv) continue;

                bh += s[i].c;
                size++;
                Union(s[i].u,s[i].v);
                if(size == n-1) break;
            }


            int flag = 0;
            for(int i =1; fibo[i] <= 100000 ; i++ )
            {
                if(fibo[i] >= bl && fibo[i] <= bh)
                {
                    flag = 1;
                    break;
                }
            }
            if(flag) printf("Yes\n");
            else printf("No\n");

        }


    }
}


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