分析：可以求簡單的任意兩點間最短距離的稍微變形，一個板子題。 

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int inf = 0x3fffff;
int gra[1005][1005];
int mon[1005][1005];
int vis[1005];
int dist[1005];
int cost[1005];
void dijkstra(int s,int d,int n)
{
    memset(vis,0,sizeof(vis));
    for(int i = 0; i <= n; i ++) dist[i] = gra[s][i],cost[i] = mon[s][i];
    int Min = inf, v, Micost = inf;
    for(int i = 1; i <= n; i ++)
    {
        Min = Micost = inf;
        for(int j = 1; j <= n; j ++)
        {
            if(!vis[j]){
                if(dist[j] < Min){Min = dist[j];
                v = j;
                Micost = cost[j];
                }
                else if(dist[j] == Min)
                {
                    if(cost[j] < Micost) {
                        Min = dist[j];
                        v = j;
                        Micost = cost[j];
                    }
                }
            }
        }
        vis[v] = 1;
        for(int j = 1; j <= n; j ++)
        {
            if(!vis[j]){
                if(dist[j] > gra[v][j] + dist[v])dist[j] = gra[v][j] + dist[v],cost[j] = mon[v][j] + cost[v];
                else if(dist[j] == gra[v][j] + dist[v]) {
                    if(cost[j] >  cost[v] + mon[v][j]){
                        dist[j] = gra[v][j] + dist[v],cost[j] = mon[v][j] + cost[v];
                    }
                }
            }
        }
    }
    printf("%d %d\n",dist[d], cost[d]);
}
int main()
{
    int n,m,u,v,w,p,s,d,t;
    scanf("%d",&t);
    while(t--){
    scanf("%d%d%d%d",&n,&m,&s,&d);
        for(int i = 0; i<= n; i ++)
        {
            for(int j = 0; j <= n; j ++)
            {
                if(i == j) gra[i][j] = 0,mon[i][j] = 0;
                else gra[i][j] = inf,mon[i][j] = inf;
            }
        }
        for(int i = 0; i < m; i ++)
        {
            scanf("%d%d%d%d",&u,&v,&w,&p);
            gra[u][v] = gra[v][u] = w;
            mon[u][v] = mon[v][u] = p;
        }
        dijkstra( s,d, n);
    }
    return 0;
}