HDU 4786 Fibonacci Tree (Minimum Spanning Tree), hdu4786fibonacci

HDU 4786 Fibonacci Tree (Minimum Spanning Tree), hdu4786fibonacci

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 5934 Accepted Submission (s): 1845

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?
(Maid 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 Input24 41 2 12 3 13 4 11 4 05 61 2 11 4 11 5 13 5 14 2 1

 

Sample OutputCase #1: YesCase #2: No

 

Source2013 Asia Chengdu Regional Contest

 

RecommendWe have carefully selected several similar problems for you: 6263 6262 6261 6260 6259 Similar to what ysy said yesterday In addition, this question is prompted directly in the question -- the black side is 0, and the white side is 1 In this case, we create a minimum spanning tree and a maximum spanning tree. If there is a Fibonacci number within the range of these two values, it indicates that the condition is met.  Simple proof:
For the Minimum Spanning Tree, you can delete an edge and add an existing edge. In this way, the weight is 1, and the boundary is the maximum spanning tree.

#include<cstdio>#include<algorithm>using namespace std;const int MAXN=1e6+10,INF=1e9+10;inline char nc(){    static char buf[MAXN],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++;}inline int read(){    char c=nc();int x=0,f=1;    while(c<'0'||c>'9'){if(c=='-')f=-1;c=nc();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=nc();}    return x*f;}struct node{    int u,v,w;}edge[MAXN];int num=1;inline void AddEdge(int x,int y,int z){    edge[num].u=x;    edge[num].v=y;    edge[num].w=z;num++;}int N,M;int fib[MAXN];int fa[MAXN];int comp1(const node &a,const node &b){return a.w<b.w;}int comp2(const node &a,const node &b){return a.w>b.w;}int find(int x){    if(fa[x]==x) return fa[x];    else return fa[x]=find(fa[x]);}void unionn(int x,int y){    int fx=find(x);    int fy=find(y);    fa[fx]=fy;}int Kruskal(int opt){    if(opt==1) sort(edge+1,edge+num,comp1);    else sort(edge+1,edge+num,comp2);    int ans=0,tot=0;    for(int i=1;i<=num-1;i++)    {        int x=edge[i].u,y=edge[i].v,z=edge[i].w;        if(find(x) == find(y)) continue;        unionn(x,y);        tot++;        ans=ans+z;        if(tot==N-1) return ans;    }}int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else    #endif    int Test=read();    fib[1]=1;fib[2]=2;    for(int i=3;i<=66;i++) fib[i]=fib[i-1]+fib[i-2];     int cnt=0;     while(Test--)    {        N=read(),M=read();num=1;        for(int i=1;i<=N;i++) fa[i]=i;        for(int i=1;i<=M;i++)        {            int x=read(),y=read(),z=read();            AddEdge(x,y,z);            AddEdge(y,x,z);        }        int minn=Kruskal(1);        for(int i=1;i<=N;i++) fa[i]=i;        int maxx=Kruskal(2);        bool flag=0;        for(int i=1;i<=66;i++)            if(minn <= fib[i] && fib[i] <= maxx)                 {printf("Case #%d: Yes\n",++cnt);flag=1;break;}        if(flag==0) printf("Case #%d: No\n",++cnt);    }    return 0;}