hdu4786Fibonacci Tree

Fibonacci Tree

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 38    Accepted Submission(s): 21

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

 

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题意：给你一些白色跟黑色的边，问你在这幅图里能不能找出一个生成树且里面的白边数量是Fibonacci数列的数

分析：找出白边最少的生成树和白边最多的生成树的情况，然后看在这之间有没有Fibonacci数，如果有就是yes啦，没有就no

为什么呢？因为生成树是n-1条边，如果你删除一条黑边，必然会孤立一个点，所以你要另一条边来连接这一个点，如果有白边就直接连这种情况就是

白边加一了，如果没有白边连出去，那可以知道这个点是没有任何的白边连到其他边。所以最多边的那副生成树也不会有这种情况，所以一定会出现中间的生成树。

还要用到并查集来判断是否在一个集合。。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int f[100005];int fa[100005];struct node{int x,y;int t;}node[100005];bool cmp(struct node a,struct node b){return a.t<b.t;}void Init(){memset(f,0,sizeof(f));int l=1,r=2,var;f[1] = f[2] = 1;while(l+r<100002){f[l+r] = 1;var = l;l = r;r = var+r;}}int find(int x){return fa[x]==x?x:(fa[x] = find(fa[x]));}int findl(int n,int m){for(int w=0;w<=n;w++)fa[w] = w;int ans = 0;for(int i=0;i<m;i++){int xx = find(node[i].x);int yy = find(node[i].y);if(xx==yy)continue;fa[xx] = yy;        ans+=node[i].t;}int z = find(1);for(int j=2;j<=n;j++)if(find(j)!=z)return 0;return ans;}int findr(int n,int m){for(int w=0;w<=n;w++)fa[w] = w;int ans = 0;for(int i=m-1;i>=0;i--){int xx = find(node[i].x);int yy = find(node[i].y);if(xx==yy)continue;fa[xx] = yy;        ans+=node[i].t;}int z = find(1);for(int j=2;j<=n;j++)if(find(j)!=z)return 0;return ans;}int main(){int n;int N,M;Init();scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d%d",&N,&M);if(M==0){printf("Case #%d: No\n",i);continue;}for(int j=0;j<M;j++){scanf("%d%d%d",&node[j].x,&node[j].y,&node[j].t);}sort(node,node+M,cmp);int l = findl(N,M);int r = findr(N,M);int FF = 0;for(int c=l;c<=r;c++)if(f[c]==1){FF = 1;break;}if(FF)printf("Case #%d: Yes\n",i);elseprintf("Case #%d: No\n",i);}return 0;}