算法模板之次短路

给你N个点和R条边，问从起点到终点的次短路是多少。无向边，且可重复走。

    #include <cstdio>    #include <cstring>    #include <queue>    #include <algorithm>    #define MAXN 100009    #define INF 0x3f3f3f3f    using namespace std;    typedef long long ll;    struct edge    {        int to, cost;        edge(int tv = 0, int tc = 0):            to(tv), cost(tc) {}    };    typedef pair<ll,ll> P;    int N, R;    vector<edge> graph[MAXN];    ll dist[MAXN];     //最短距离    ll dist2[MAXN];    //次短距离    void solve()    {        memset(dist, 0x3f, sizeof(dist));        memset(dist2, 0x3f, sizeof(dist2));        priority_queue<P, vector<P>, greater<P> > Q;        dist[1] = 0;        Q.push(P(0, 1));        while(!Q.empty())        {            P p = Q.top();            Q.pop();            int v = p.second;            ll d = p.first;            if(dist2[v] < d) continue;            for(unsigned i = 0; i < graph[v].size(); i++)            {                edge &e = graph[v][i];                ll d2 = d + e.cost;                if(dist[e.to] > d2)                {                    swap(dist[e.to], d2);                    Q.push(P(dist[e.to], e.to));                }                if(dist2[e.to] > d2 && dist[e.to] < d2)                {                    dist2[e.to] = d2;                    Q.push(P(dist2[e.to], e.to));                }            }        }        printf("%lld\n", dist2[N]);    }    int main()    {        int A, B, D;        int t;        scanf("%d",&t);        while(t--)        {            scanf("%d%d", &N, &R);            for(int i=0;i<=N;i++)                graph[i].clear();            for(int i = 0; i < R; i++)            {                scanf("%d%d%d", &A, &B, &D);                graph[A].push_back(edge(B, D));                graph[B].push_back(edge(A, D));            }            solve();        }        return 0;    }    /*    Sample Input    2    3 3    1 2 1    2 3 4    1 3 3    2 1    1 2 1    Sample Output    5    3    */