hdoj 4786 Fibonacci Tree 【最小生成树 ＋ 最大生成树】

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2563    Accepted Submission(s): 817

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

本题说的是：连通所有的点，且连通白边的数目须是fibonacci数列里面的数（可以有黑边，比如说黑3，白1就满足。并不是非得全是白边）；

 

思路很简单：先按白——>黑排序，计算连通需要的最多的白边数目max；然后黑——>白排序，计算连通需要的最少的白边数目min；最后判断是否连通以及是否存在fibonacci数列里面的数fib[ i ]满足 min>= fib[ i ] <=max 。

 

ac代码：

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define M 100000+1
using namespace std;
int set[M],fib[M];
struct line
{
    int start;
    int end;
    int judge;
}num[M];
bool cmp1(line a,line b)
{
    return a.judge>b.judge;//white放前面 
}
bool cmp2(line a,line b)
{
    return a.judge<b.judge;
}
void dabiao()
{
    int i,j;
    int t,sum;
    fib[1]=1;fib[2]=1;
    for(i=3;fib[i]<M;i++)
    {
        fib[i]=fib[i-1]+fib[i-2];
    }
}
int find(int p)
{
    int child=p;
    int t;
    while(p!=set[p])
    p=set[p];
    while(child!=p)
    {
        t=set[child];
        set[child]=p;
        child=t;
    }
    return p;
}
void merge(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    set[fx]=fy;
}
int main()
{
    int t,n,m,i,j;
    int max,min;
    int k=1,exist,sum;
    memset(fib,0,sizeof(fib));
    dabiao();//fib[i]存储fibonacci数 
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
        set[i]=i;
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&num[i].start,&num[i].end,&num[i].judge);
        }
        sort(num,num+m,cmp1);//white first
        max=0;
        for(i=0;i<m;i++)
        {
            if(find(num[i].start)!=find(num[i].end))
            {
                merge(num[i].start,num[i].end);
                if(num[i].judge)//white
                max++;
            }
        }
        min=0;
        sort(num,num+m,cmp2);//black first
        for(i=1;i<=n;i++)
        set[i]=i;
        for(i=0;i<m;i++)
        {
            if(find(num[i].start)!=find(num[i].end))
            {
                merge(num[i].start,num[i].end);
                if(num[i].judge)//white
                min++;
            }
        }
        exist=0;
        for(i=1;i<=n;i++)
        {
            if(set[i]==i)
            exist++;
            if(exist>1)
            break;
        }
        printf("Case #%d: ",k++);
        if(exist>1)//can not 
        {
            printf("No\n");
            continue;
        }
        exist=0;
        for(i=0;fib[i]<max;i++)
        {
            if(fib[i]>=min&&fib[i]<=max)
            {
                exist=1;
                break;
            }
        }
        if(exist)
        printf("Yes\n");
        else
        printf("No\n");
    }
    return 0;
}