http://codeforces.com/gym/101061/problem/C
题意:有n个十字路口 , m条双向道路,步行道和车道,问从u到v,要使步行路程尽可能的小,如果步行路程相同,则总路程尽可能小。
解法:以步行为第一优先级,车程为第二优先级,有车道乘车则步行就赋值为0,如果有步行无车道则该车道赋值为0。
//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#include <stdlib.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 100
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e5+9;
const int maxn = 1e2+9;
const double esp = 1e-6;
int n , m ;
int c[maxn][maxn] , p[maxn][maxn] , vis[maxn] , dis[maxn];
int val[maxn];
void dijkstra(int u){
rep(i , 1 , n){
dis[i] = p[u][i];
val[i] = c[u][i];
}
vis[u] = 1 ;
rep(i , 1 , n-1){
int pos = -1;
int map = INF , mac = INF;
rep(j , 1 , n){
if(!vis[j] && map > dis[j]){
map = dis[j];
mac = val[j];
pos = j ;
}else if(!vis[j] && map == dis[j] && mac > val[j]){
mac = val[j];
pos = j ;
}
}
if(pos == -1) break;
vis[pos] = 1 ;
rep(j , 1 , n){
if(!vis[j] && dis[j] > dis[pos] + p[pos][j]){
dis[j] = dis[pos] + p[pos][j];
val[j] = val[pos] + c[pos][j];
}else if(!vis[j] && dis[j] == dis[pos] + p[pos][j] && val[j] > val[pos] + c[pos][j]){
val[j] = val[pos] + c[pos][j];
}
}
}
}
void init(){
ME(vis , 0);
fill(p[0] , p[0]+maxn*maxn , INF);
fill(c[0] , c[0]+maxn*maxn , INF);
}
void solve(){
init();
scanf("%lld%lld" , &n , &m);
rep(i , 1 , m){
int u , v , w , k ;
scanf("%lld%lld%lld%lld" , &u , &v , &w ,&k);
if(k == 1){
p[u][v] = p[v][u] = min(p[u][v] , w);
}else{
c[u][v] = c[v][u] = min(c[u][v] , w);
}
}
rep(i , 1 , n){
rep(j , 1 , n){
if(c[i][j] != INF){
p[i][j] = 0 ;
}else if(c[i][j] == INF && p[i][j] != INF){
c[i][j] = 0 ;
}
}
}
int u , v ;
scanf("%lld%lld" , &u , &v);
dijkstra(u);
if(dis[v] == INF && val[v] == INF) cout << -1 << endl;
else cout << dis[v] << " " << dis[v]+val[v] << endl;
}
signed main()
{
int t ;
cin >> t ;
while(t--){
solve();
}
}