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, ... )
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).
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
Recommend
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);
else
printf("Case #%d: No\n",i);
}
return 0;
}
作者：zhangmin4235 发表于2013-11-17 16:15:59 原文链接
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