HDU 4786 Fibonacci Tree

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4786

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 <= 10 ⁵) and M(0 <= M <= 10 ⁵). 
　　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

/*
题意：
给你一个图，每条边有两种颜色黑色或者白色，让你判断存不存在一棵生成树，使得白边的数量为斐波那契数。
分别求出白边在生成树中的最大值和最小值，用生成树思想来做。
当然如果图不是连通图的话，则肯定输出No。然后判断在这个最大最小之间存不存斐波那契数，如果存在则Yes，否则No。因为在这个区间内总能找到一条白边可以用黑边来代替。


这题我刚开始用的是Prim算法，但是这题边数有100000条，就TLE了，难过。。
后来改用kruskal，马上就过了。。诶
*/
#include <iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int team[100005],a[30];
struct node
{
    int x,y,z;
}edge[100005];
int n,m,summax,summin,t,len;

bool cmp1(node a,node b) //从小到大排列
{
    return a.z<b.z;
}
bool cmp2(node a,node b)//从大到小排列
{
    return a.z>b.z;
}
int findteam(int k)//并查集
{
    if (team[k]==k) return k;
     else
     {
         team[k]=findteam(team[k]);
         return team[k];
     }
}

void primmin()//求最小生成树，求summin
{
    sort(edge+1,edge+m+1,cmp1);
    for(int i=1;i<=m;i++)
    {
        int teamx=findteam(edge[i].x);
        int teamy=findteam(edge[i].y);
        if (teamx==teamy) continue;
        summin+=edge[i].z;
        team[teamx]=teamy;
    }
}
void primmax()//求最大生成树，求summax
{
    sort(edge+1,edge+m+1,cmp2);
    for(int i=1;i<=m;i++)
    {
        int teamx=findteam(edge[i].x);
        int teamy=findteam(edge[i].y);
        if (teamx==teamy) continue;
        summax+=edge[i].z;
        team[teamx]=teamy;
    }
}
void prework()//把小于10^5次的菲波那切数列求出来
{
    len=2;
    a[1]=a[2]=1;
    for(len=3;a[len-1]<100000;len++) a[len]=a[len-1]+a[len-2];
    len--;
}
bool ok()//判断能不能连通
{
    int x=findteam(team[1]);
    for(int i=2;i<=n;i++)
    {
        int y=findteam(team[i]);
        if (x!=y) {printf("No
"); return 0;}
    }
    return 1;
}

int main()
{
    prework();
    scanf("%d",&t);
    for(int ii=1;ii<=t;ii++)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
            scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].z);

        printf("Case #%d: ",ii);
        summin=summax=0;
        for(int i=1;i<=n;i++) team[i]=i;
        primmin();
        if (!ok()) continue;
        for(int i=1;i<=n;i++) team[i]=i;
        primmax();
        if (!ok()) continue;

        for(int i=1;i<=len;i++)
        {
            if (a[i]>=summin && a[i]<=summax)  {printf("Yes
");break;}
            if (a[i]>summax) {printf("No
");break;}
        }
    }
    return 0;
}