最短路 || POJ 1511 Invitation Cards

已知图中从一点到另一点的距离，从1号点到另一点再从这一点返回1号点，求去到所有点的距离之和最小值

*解法：正着反着分别建图，把到每个点的距离加起来

spfa跑完之后dist数组就是从起点到每一点的最短距离

get反着建图这种技能

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define SZ 1000005
#define INF 1e18+10
long long dist[SZ], tmp[SZ], dis = ;
int head[SZ], nxt[SZ], tot = ;
int x[SZ], y[SZ];
long long z[SZ];
int use[SZ];
struct edge
{
    int t;
    long long d;
}l[SZ];
void build(int f, int t, long long d)
{
    l[++tot] = (edge){t, d};
    nxt[tot] = head[f];
    head[f] = tot;
}
queue<int> q;
int p, qq;
void spfa()
{
    for(int i = ; i <= p; i++) dist[i] = INF, use[i] = ;
    use[] = , dist[] = ;
    q.push();
    while(q.size())
    {
        int u = q.front();
        q.pop();
        use[u] = ;
        for(int i = head[u]; i; i = nxt[i])
        {
            int v = l[i].t;
            if(dist[v] > dist[u] + l[i].d)
            {
                dist[v] = dist[u] + l[i].d;
                if(!use[v])
                    use[v] = , q.push(v);
            }
        }
    }
}
int main()
{
    int T, s = ;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d", &p, &qq);
        dis = ;
        tot = ;
        for(int i = ; i < qq; i++) head[i] = nxt[i] = use[i] = ;
        for(int i = ; i < qq; i++)
        {
            scanf("%d %d %lld", &x[i], &y[i], &z[i]);
            build(x[i], y[i], z[i]);
        }
        spfa();
        for(int i = ; i <= p; i++) tmp[i] = dist[i];
        tot = ;
        for(int i = ; i < qq; i++)
            head[i] = nxt[i] = use[i] = ;
        for(int i = ; i < qq; i++)
            build(y[i], x[i], z[i]);
        spfa();
        for(int i = ; i <= p; i++)
            dis = dis + tmp[i] + dist[i];
        printf("%lld\n", dis);
    }
    return ;
}