Fibonacci Tree 最小生成树

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 434    Accepted Submission(s): 146

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

 

 

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const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 300000;
vector<int> g[maxn];
struct Edge
{
    int a;
    int b;
    int c;
    Edge(int _a,int _b,int _c):a(_a),b(_b),c(_c){}
};
void init()
{
    repf(i,1,100010)
        g[i].clear();
}
vector<Edge> edges;
int p[maxn];
int find(int x)
{
    if(p[x] == x)
        return x;
    else
    {
        p[x] = find(p[x]);
        return p[x];
    }
}

int kruskal()
{
    int ans = 0;
    rep(i,0,edges.size())
    {
        int x = find(edges[i].a);
        int y = find(edges[i].b);
        if(x != y)
        {
            p[x] = y;
            ans += edges[i].c;
        }
    }
    return ans;
}

bool cmp1(Edge e1,Edge e2)
{
    return e1.c > e2.c;
}
bool cmp2(Edge e1,Edge e2)
{
    return e1.c < e2.c;
}

int vis[maxn];

int vis1[maxn];
void dfs(int x)
{
    if(vis1[x]) return ;
    vis1[x] = 1;
    rep(i,0,g[x].size())
        dfs(g[x][i]);
}
int main() 
{
    //freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    clr(vis);
    vis[1] = 1;
    int a = 1;
    int b = 1;
    while(a + b <= 100000)
    {
        vis[a + b] = 1;
        int t = a;
        a = b;
        b = t + b;
    }
    int k = 1;
    while(T--)
    {
        edges.clear();
        init();
        clr(vis1);
        int n,m;
        scanf("%d%d",&n,&m);
        repf(i,1,n) p[i] = i;
        repf(i,1,m)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            g[a].push_back(b);
            g[b].push_back(a);
            edges.push_back(Edge(a,b,c));
        }
        dfs(1);
        int tag = 1;
        repf(i,1,n) if(vis1[i] == 0) 
        {
            tag = 0;break;
        }
        if(tag == 0)
        {
            printf("Case #%d: ",k++);
            cout<<"No"<<endl;
            continue;
        }
        sort(edges.begin(),edges.end(),cmp1);
        
        int  Max = kruskal();
        repf(i,1,n) p[i] = i;
        sort(edges.begin(),edges.end(),cmp2);
        
        int Min = kruskal();
        
        int flag = 0;
        repf(i,Min,Max)
            if(vis[i])
            {
                flag =1;
                break;
            }
        printf("Case #%d: ",k++);
        if(flag)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}