desc cpp using tran soft fine put sea can

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4786

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

題意：

N個頂點,M條邊。每條邊或為白色或為黑色（ 1 or 0 ）。

問有沒實用是斐波那契數的數目的白色邊構成一棵生成樹。

PS:

事實上說是並查集更靠譜一點的醬紫！

首先推斷整個圖是否是連通的，若不連通則直接輸出No。

接下來首先僅討論白邊。不要黑邊，看最多能增加多少條白邊。使得不存在環。

這樣我們得到了能增加白邊的最大值max。（就是全部生成樹裏白邊數量的最大值）。

接下來同理僅討論黑邊，這樣我們能夠得到可增加白邊的最小值min。（也能夠覺得是全部生成樹中白邊的最小值）。

然後我們僅僅要推斷這兩個值之間是否存在斐波那契數即可了。

代碼例如以下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 100017;
struct node
{
    int u, v;
    int c;
} a[maxn];
int f[maxn], Fib[maxn];
int n, m;
int findd(int x)
{
    return x==f[x] ? x : f[x]=findd(f[x]);
}
int kruskal(int sign)
{
    int k = 0;
    //sort(a,a+m,cmp);
    for(int i = 0; i <= n; i++)
    {
        f[i] = i;
    }
    for(int i = 1; i <= m; i++)
    {
        if(a[i].c != sign)
        {
            int f1 = findd(a[i].u);
            int f2 = findd(a[i].v);
            if(f1 != f2)
            {
                f[f1] = f2;
                k++;
            }
        }
    }
    return k;
}
void init()
{
    Fib[0] = 1, Fib[1] = 2;
    for(int i = 2; ; i++)
    {
        Fib[i] = Fib[i-1]+Fib[i-2];
        if(Fib[i] > maxn)
            break;
    }
}
int main()
{
    int t;
    int cas = 0;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].c);
        }
        int ans = kruskal(2);
        if(ans != n-1)//不能形成樹
        {
            printf("Case #%d: No\n",++cas);
            continue;
        }
        int maxx = kruskal(0);
        int minn = n-1-kruskal(1);
        int flag = 0;
        for(int i = 0; ; i++)
        {
            if(Fib[i] >=minn && Fib[i]<=maxx)
            {
                flag = 1;
                break;
            }
            if(Fib[i] > maxx)
            {
                break;
            }
        }
        if(flag)
        {
            printf("Case #%d: Yes\n",++cas);
        }
        else
        {
            printf("Case #%d: No\n",++cas);
        }
    }
    return 0;
}

HDU 4786（最小生成樹 kruskal）