杭電4786--Fibonacci Tree（生成樹）

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

題意: 存不存在所含白邊數為菲波那契數的生成樹（包含圖中所有節點）。

　　　找出生成樹中所含白邊最多和最少的生成樹中所含白邊數， 就是按照邊的黑白排一下序， 組成一個區間[a,  b]； 然後在區間[a, b]中枚舉菲波那契數。 參考bin神代碼。 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; 
bool flag1, flag2;  int sum, Y;
struct Sorrow
{
    int a, b, c;
} edge[100001];
bool cmp1(Sorrow a, Sorrow b)
{
    return a.c < b.c;
}
bool cmp2(Sorrow a, Sorrow b)
{
    return a.c > b.c;
}
int father[100001], n;
void init()
{
    //memset(father, -1, sizeof(father));
    for(int i = 1; i <= n; i++)
        father[i] = i; 
}
int Find(int a)
{
    if(father[a] == a)
        return a;
    else
        return father[a] = Find(father[a]);
} 
int mercy(int a, int b, int c)
{
//    int Y = 0;
    int Q = Find(a);
    int P = Find(b);
    if(Q != P)
    {
        father[Q] = P;
        if(c)
            Y++;
    } 
    return Y;
}
bool Judge()
{
    int cnt = 0;
    for(int i = 1; i <= n; i++)
        if(father[i] == i)
        {
            cnt++;
            if(cnt > 1)
                return false;
        }
    return true;
}
int b[100001]; 
int Bitch()
{
    sum = 1;
    b[0] = 1; b[1] = 2;
    while(b[sum] < 100001)
    {
        b[sum+1] = b[sum] + b[sum-1];
        sum++;  
    }
}
int main()
{
    int T, Q = 1;
    scanf("%d", &T);
    while(T--)
    {
        int m, L, R;
         
        scanf("%d %d", &n, &m);  init(); 
        for(int i = 0; i < m; i++)
            scanf("%d%d%d", &edge[i].a, &edge[i].b, &edge[i].c);
        sort(edge, edge+m, cmp1); Y = 0;
        for(int i = 0; i < m; i++)
            L = mercy(edge[i].a, edge[i].b, edge[i].c);
       // flag1 = Judge();
        init();
        sort(edge, edge+m, cmp2); Y = 0;
        for(int i = 0; i < m; i++)
            R = mercy(edge[i].a, edge[i].b, edge[i].c);
        flag2 = Judge();
        printf("Case #%d: ", Q++);
        if(flag2 == false)
        {
            printf("No\n"); 
            continue;
        }
        Bitch();
        bool flag = false;
        for(int i = 0; i < sum; i++)
        {
            if(b[i] >= L && b[i] <= R)
                flag = true;
        } 
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;    
}      

轉載于:https://www.cnblogs.com/soTired/p/4827753.html