一.题目链接：

ZOJ-3946

二.题目大意：

T 组数据.

第一行两个整数 n，m   

之后 m 行数据，每行给出第 i 条路的 {起点，终点，花费时间，花费金钱}

首都为第 0 号城市.

求从首都到其他所有城市所需的 总时间 和 总建路花费.

三.分析：

双权值的单源最短路，更改 if 条件语句里就可以了.

注意：时间可重复加，但花费不可以.

所以直接用 dis2[v] 来记录第 i 条路的花费

之后求和即可.

还有要开 long long int.

详见代码.

四.分析：

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-6
#define pi acos(-1.0)
#define ll long long int
using namespace std;

const int M = (int)1e5;
const ll inf = 1ll << 50;

struct node
{
    int u;
    int v;
    int c1;
    int c2;
    int next;
}Edge[2 * M + 5];

int cnt = 0;
bool vis[M + 5];
int head[M + 5];
ll dis1[M + 5];
ll dis2[M + 5];

void init(int n)
{
    cnt = 0;
    for(int i = 0; i < n; ++i)
    {
        vis[i] = 0;
        head[i] = -1;
        dis1[i] = inf;
        dis2[i] = inf;
    }
}

void add(int u, int v, int c1, int c2)
{
    Edge[cnt].u = u;
    Edge[cnt].v = v;
    Edge[cnt].c1 = c1;
    Edge[cnt].c2 = c2;
    Edge[cnt].next = head[u];
    head[u] = cnt++;
}

struct cmp
{
    bool operator()(int a, int b)
    {
        if(dis1[a] == dis1[b])
            return dis2[a] > dis2[b];
        return dis1[a] > dis1[b];
    }
};

void spfa(int start)
{
    vis[start] = 1;
    dis1[start] = 0;
    dis2[start] = 0;
    priority_queue <int, vector <int>, cmp> q;
    q.push(start);
    while(!q.empty())
    {
        ll u = q.top();
        q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = Edge[i].next)
        {
            int v = Edge[i].v;
            if(dis1[v] >= dis1[u] + Edge[i].c1)
            {
                if(dis1[v] == dis1[u] + Edge[i].c1)
                    dis2[v] = min(dis2[v], 1ll * Edge[i].c2);
                else
                    dis2[v] = 1ll * Edge[i].c2;
                dis1[v] = dis1[u] + Edge[i].c1;
                if(!vis[v])
                {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n, m;
        scanf("%d %d", &n , &m);
        init(n);
        for(ll i = 0; i < m; ++i)
        {
            ll u, v, c1, c2;
            scanf("%d %d %d %d", &u, &v, &c1, &c2);
            add(u, v, c1, c2);
            add(v, u, c1, c2);
        }
        spfa(0);
        ll sum1 = 0;
        ll sum2 = 0;
        for(int i = 0; i < n; ++i)
        {
            sum1 += dis1[i];
            sum2 += dis2[i];
        }
        printf("%lld %lld\n", sum1, sum2);
    }
    return 0;
}