HDU ~ 2066 ~ 一个人的旅行 （反向建边 or 建超级源 + 最短路）

思路

同HDU ~ 2680 ~ Choose the best route （反向建边 or 超级源点 + 最短路）

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e3 + 5;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int from, to, dist;       //起点,终点,距离
    Edge(int u, int v, int w):from(u), to(v), dist(w) {}
};

struct Dijkstra
{
    int n, m;                 //结点数,边数(包括反向弧)
    vector<Edge> edges;       //边表。edges[e]和edges[e^1]互为反向弧
    vector<int> G[MAXN];      //邻接表，G[i][j]表示结点i的第j条边在edges数组中的序号
    int vis[MAXN];            //标记数组
    int d[MAXN];              //s到各个点的最短路
    int pre[MAXN];            //上一条弧

    void init(int n)
    {
        this->n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].clear();
    }

    void add_edge(int from, int to, int dist)
    {
        edges.push_back(Edge(from, to, dist));
        m = edges.size();
        G[from].push_back(m - 1);
    }

    struct HeapNode
    {
        int from; int dist;
        bool operator < (const HeapNode& rhs) const
        {
            return rhs.dist < dist;
        }
        HeapNode(int u, int w): from(u), dist(w) {}
    };

    void dijkstra(int s)
    {
        priority_queue<HeapNode> Q;
        for (int i = 0; i <= n; i++) d[i] = INF;
        memset(vis, 0, sizeof(vis));
        d[s] = 0;
        Q.push(HeapNode(s, 0));
        while (!Q.empty())
        {
            HeapNode x = Q.top(); Q.pop();
            int u = x.from;
            if (vis[u]) continue;
            vis[u] = true;
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge& e = edges[G[u][i]];
                if (d[e.to] > d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    pre[e.to] = G[u][i];
                    Q.push(HeapNode(e.to, d[e.to]));
                }
            }
        }
    }
};

int T, S, D;
Dijkstra solve;

int main()
{
    while(~scanf("%d%d%d", &T, &S, &D))
    {
        solve.init(MAXN);
        while(T--)
        {
            int a, b, time;
            scanf("%d%d%d", &a, &b, &time);
            solve.add_edge(a, b, time);
            solve.add_edge(b, a, time);
        }
        while (S--)
        {
            int a; scanf("%d", &a);
            solve.add_edge(0, a, 0);
        }
        solve.dijkstra(0);
        int MIN = INF;
        while (D--)
        {
            int a; scanf("%d", &a);
            MIN = min(MIN, solve.d[a]);
        }
        printf("%d\n", MIN);
    }
    return 0;
}
/*
6 2 3
1 3 5
1 4 7
2 8 12
3 8 4
4 9 12
9 10 2
1 2
8 9 10
*/