2013成都站F题||hdu4786 并查集 生成树
http://acm.hdu.edu.cn/showproblem.php?pid=4786
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 <= 10 5) and M(0 <= M <= 10 5).
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 <= 10 5) and M(0 <= M <= 10 5).
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
解题思路：
按边的颜色排序两次，一次黑边优先，利用并查集构造一个生成树，此时是所有方法中白边最少的情况，x。而后白边优先，得出的是白边最多的情况，y。找找x，y二者之间是否包括一个Fibonacci数即可。
#include <string.h>
#include <stdio.h>
#include <iostream>
#include <algorithm>
//#define debug
using namespace std;
struct note
{
int u,v,c;
} edge[100005];
int n,m,fa[100005],num;
bool cmp1(note a,note b)
{
return a.c<b.c;
}
bool cmp2(note a,note b)
{
return a.c>b.c;
}
void initset()
{
for(int i=0; i<=n; i++)
fa[i]=i;
}
int find(int x)
{
if(x==fa[x])
return x;
return fa[x]=find(fa[x]);
}
int un(int x,int y)
{
x=find(x);
y=find(y);
if(x!=y)
{
fa[y]=x;
num++;
return 1;
}
return 0;
}
int a[1005];
void fib()
{
a[0]=1;
a[1]=1;
for(int i=2; i<30; i++)
a[i]=a[i-1]+a[i-2];
#ifdef debug
printf("%d\n",a[29]);
#endif // debug
}
int main()
{
int T,tt=0;
scanf("%d",&T);
fib();
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0; i<m; i++)
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].c);
initset();
sort(edge,edge+m,cmp1);
int cnt1;
num=0;
int count=0;
for(int i=0; i<m; i++)
{
if(un(edge[i].u,edge[i].v))
count+=edge[i].c;
if(num==n-1)
{
cnt1=count;
break;
}
}
initset();
sort(edge,edge+m,cmp2);
int cnt2;
num=0;
count=0;
for(int i=0; i<m; i++)
{
if(un(edge[i].u,edge[i].v))
count+=edge[i].c;
if(num==n-1)
{
cnt2=count;
break;
}
}
int flag=0;
for(int i=0; i<29; i++)
if(cnt1==a[i]||cnt2==a[i])
{
printf("Case #%d: Yes\n",++tt);
flag=1;
break;
}
if(flag)
continue;
int x=upper_bound(a,a+29,cnt1)-a;
int y=upper_bound(a,a+29,cnt2)-a;
if(x==y)
printf("Case #%d: No\n",++tt);
else
printf("Case #%d: Yes\n",++tt);
}
return 0;
}