HDU

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5844    Accepted Submission(s): 1823

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

24 41 2 12 3 13 4 11 4 05 61 2 11 3 11 4 11 5 13 5 14 2 1

 

Sample Output

Case #1: YesCase #2: No

 

Source

2013 Asia Chengdu Regional Contest 

 

题意：无向图有N个点M条边，每条边有黑或白两种颜色，求该图是否存在一个生成树使得生成树中白边的数量为斐波那契数

白边权值为1，黑边为0，求出生成树中最少有多少条白边和最多有多少条白边，mi，mx

若区间[mi,mx]间存在斐波那契数，则必定存在一个生成树其中含有的白边数量为斐波那契数

当mi < mx 时，必定有存在(mx-mi)条黑边，将任一条黑边删去后所留下的两个连通块一定可以用一条白边连接起来

问题就变成了求最大与最小生成树的权值， 注意判断是否能够形成生成树

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int N = 1e5 +10;int t,n,m,cas,fib[N],fa[N];int mi, mx;struct node {    int from, to, cost;}e[N];bool cmp1(node a,node b){return a.cost<b.cost;}bool cmp2(node a,node b){return a.cost>b.cost;}int find(int x){    int be = x;    while(fa[x]!=x) x = fa[x];    while(fa[be]!=be){        int c = fa[be];        fa[be] = x;        be = c;    }    return x;}int slove(){    int cnt = 0, ans = 0;    for(int i=0;i<m;i++){        int fx = find(e[i].from), fy = find(e[i].to);        if(fx == fy) continue;        fa[fx] = fy; cnt++; ans += e[i].cost;        if(cnt==n-1) return ans;    }    return n+1;}int main(){    fib[1] = 1;    for(int a = 1,b = 2;b<N;){        fib[b] = 1;        int c = b;        b = a + b;        a = c;    }    for(scanf("%d",&t),cas = 0;t--;){        scanf("%d%d",&n,&m);        printf("Case #%d: ",++cas);        mx = 0; mi = n;        for(int i=0;i<m;i++){            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            e[i].from = x; e[i].to = y; e[i].cost = z;        }        for(int i=1;i<=n;i++) fa[i] = i;        sort(e,e+m,cmp1);        mi = slove();        if(mi>=n){            printf("No\n");            continue;        }        for(int i=1;i<=n;i++) fa[i] = i;        sort(e,e+m,cmp2);        mx = slove();        bool flag = false;        for(int i=mi;i<=mx;i++){            if(fib[i]) {                flag = true; break;            }        }                if(flag) printf("Yes\n");        else printf("No\n");    }    return 0;}