HDU - 4786 Fibonacci Tree 最小生成树+最大生成树

题目链接<-点击

　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

没想到这题居然是区域赛的题，应该是一个签到题吧

题目大意，就是有两种颜色的边，黑边和白边，你要把所有的点连接起来，并使得你用的白边树是斐波那契数列中的一个数。

思路：

先打表斐波那契，然后求出来最大生成树（尽可能使用白边）和最小生成树（尽可能使用黑边），若最大值和最小值之间有斐波那契数列中的一个数，就正确，否则就没有。

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1000000+100;
int vis[maxn],post[maxn];
int pre[maxn];
int n,m;
struct node
{
int from,to,val;
}road[maxn];
int cmp(node aa,node bb)
{
return aa.val>bb.val;
}
int cmp1(node aa,node bb)
{
return aa.val<bb.val;
}
int findd(int x)
{
int r=x;
while(r!=pre[r])
r=p
4000
re[r];
int i=x,j;
while(i!=r)
{
j=pre[i];
pre[i]=r;
i=j;
}
return r;
}
int zyz()
{
int i;
for(i=0;i<=n;i++)
{
pre[i]=i;
}
sort(road+1,road+1+m,cmp);
int ans=0;
int aaaa=1;
for(i=1;i<=m;i++)
{
int tx=findd(road[i].from);
int ty=findd(road[i].to);
if(tx!=ty)
{
pre[ty]=tx;
ans+=road[i].val;
aaaa++;
}
}
if(aaaa<n)
return -1;
else
return ans;
}
int zhang()
{
int i;
for(i=0;i<=n;i++)
{
pre[i]=i;
}
sort(road+1,road+1+m,cmp1);
int ans=0;
int aaaa=1;
for(i=1;i<=m;i++)
{
int tx=findd(road[i].from);
int ty=findd(road[i].to);
if(tx!=ty)
{
pre[ty]=tx;
ans+=road[i].val;
aaaa++;
}
}
if(aaaa<n)//注意啊！！！！wa了四发，这里以前从来没注意过！！！
return -1;
else
return ans;
}
int main ()
{
int t;
memset(post,0,sizeof(post));
memset(vis,0,sizeof(vis));
vis[1]=1;
vis[2]=1;
post[0]=1;
post[1]=2;
int cnt=1;
while(post[cnt]<maxn)
{
post[++cnt]=post[cnt-1]+post[cnt-2];
if(post[cnt]<maxn)
vis[post[cnt]]=1;
}
scanf("%d",&t);
int kk;
for(kk=1;kk<=t;kk++)
{
scanf("%d%d",&n,&m);
int i;
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&road[i].from,&road[i].to,&road[i].val);
}
int ed=zyz();
int st=zhang();
int flag=0;
if(ed==-1||st==-1)
{
printf("Case #%d: No\n",kk);
continue;
}
for(i=st;i<=ed;i++)
{
if(vis[i])
{
flag=1;
break;
}
}
if(flag)
printf("Case #%d: Yes\n",kk);
else
printf("Case #%d: No\n",kk);
}
}