Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
One day Petya encountered a tree with n vertexes. Besides, the tree was weighted, i. e. each edge of the tree has weight (a positive integer). An edge is lucky if its weight is a lucky number. Note that a tree with n vertexes is an undirected connected graph that has exactly n - 1 edges.
Petya wondered how many vertex triples (i, j, k) exists that on the way from i to j, as well as on the way from i to k there must be at least one lucky edge (all three vertexes are pairwise distinct). The order of numbers in the triple matters, that is, the triple (1, 2, 3) is not equal to the triple (2, 1, 3) and is not equal to the triple (1, 3, 2).
Find how many such triples of vertexes exist.
Input
The first line contains the single integer n (1 ≤ n ≤ 105) — the number of tree vertexes. Next n - 1 lines contain three integers each: ui vi wi (1 ≤ ui, vi ≤ n, 1 ≤ wi ≤ 109) — the pair of vertexes connected by the edge and the edge‘s weight.
Output
On the single line print the single number — the answer.
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is recommended to use the cin, cout streams or the %I64d specificator.
Sample test(s)
input
41 2 43 1 21 4 7
output
16
input
41 2 41 3 471 4 7447
output
24
Note
The 16 triples of vertexes from the first sample are:(1, 2, 4), (1, 4, 2), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 3, 4), (2, 4, 1), (2, 4, 3), (3, 2, 4), (3, 4, 2), (4, 1, 2), (4, 1, 3), (4, 2, 1), (4, 2, 3), (4, 3, 1), (4, 3, 2).
In the second sample all the triples should be counted: 4·3·2 = 24.
题意：
一条边，如果这一条边的权值只由数字4,7组成，我们就说这条边是一条幸运边，否则不是
现在有一棵树，边有权值，问:
这棵树有多少点对(i,j,k)满足：
i!=j!=k,
并且路径i到j至少经过一条幸运边，路径i到k也至少经过一条幸运边
注意：
点对(i,j,k)和(j,i,k)和(i,k,j)算是不同的点对
这道题，直接算满足的点对好像有点麻烦，正难则反，我们先算出不符合的点对
符合的点对=总点对-不符合的点对
总点对=n*(n-1)*(n-2)
现在考虑不符合的点对
不符合的点对分2种：
1.i到j和i到k2条路径都不包含幸运边
2.2条路径，其中一条经过了幸运边，另外一条没有经过幸运边
这棵树，根据幸运边我们可以分成若干个联通块，这里可以用并查集来实现
find_fa()函数的同时维护数组sum
sum[i]表示i所在的联通块的节点数
接着我们枚举联通块
对于每一个连通块:
2条路径都没有经过幸运边的数量：sum[i]*(sum[i]-1)*(sum[i]-2)
只有一条经过幸运边的数量：2*sum[i]*(sum[i]-1)*(n-sum[i])
累加就得到所有不符合的点对了
注意：
1.并查集记得路径压缩
2.每一个连通块只算一次，不要重复计算
代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
using namespace std;
const int maxn=1e5+5;
int fa[maxn];
LL sum[maxn];
void solve();
int main()
{
solve();
return 0;
}
//初始化
void init(int n)
{
for(int i=1;i<=n;i++){
fa[i]=i;
sum[i]=1;
}
}
//判断是不是幸运边
bool ok(int w)
{
while(w){
int cur=w%10;
if(cur!=4 && cur!=7)
return false;
w/=10;
}
return true;
}
//并查集，注意要路径压缩，注意sum数组的更新
int find_fa(int x)
{
if(fa[x]==x)
return x;
int cur=find_fa(fa[x]);
sum[cur]+=sum[x];
sum[x]=0;
fa[x]=cur;
return cur;
}
void solve()
{
int n;
scanf("%d",&n);
init(n);
for(int i=1;i<n;i++){
int u,v,w;
scanf("%d %d %d",&u,&v,&w);
if(!ok(w)){
int fau=find_fa(u);
int fav=find_fa(v);
if(fau!=fav){
fa[fau]=fav;
sum[fav]+=sum[fau];
sum[fau]=0;
}
}
}
if(n<3){
printf("0\n");
return ;
}
LL ans=0;
for(int i=1;i<=n;i++){
if(find_fa(i)==i){
if(sum[i]>2)
ans+=(sum[i]*(sum[i]-1)*(sum[i]-2));
ans+=2LL*sum[i]*(sum[i]-1)*(n-sum[i]);
}
}
ans=n*(n-1LL)*(n-2LL)-ans;
printf("%I64d\n",ans);
return ;
}