PAT---1030 Travel Plan(迪杰斯特拉，记录最短路径)

题意：输出最短路径，若最短路径相同，则输出代价最小的路径。
思路：迪杰斯特拉算法，用path数组记录最短路径，path[a] = b,表示从源s到b的最短路径中，最后一跳是通过a到b的。故输出时，应该用栈将b入栈后，依次将path[]数组从后往前遍历进栈。然后输出栈即可。

#include <iostream>
#include <vector>
#include <cstring>
#include <stack>

using namespace std;

struct Node
{
    int next, dis, cost;
};

vector<Node> edges[501];
bool visited[501];
int dist[501];
int cost[501];     //记录花费
int path[501];    //记录路径

void print(int d)
{
    stack<int> S;
    S.push(d);
    int pre = path[d];
    while(pre != -1)
    {
        S.push(pre);
        pre = path[pre];
    }
    while(!S.empty())
    {
        cout << S.top() << " ";
        S.pop();
    }
}

void Dij(int n, int s)
{
    memset(visited, 0 , sizeof(visited));
    for(int i = 0; i < n; i++)
    {
        dist[i] = -1;
        cost[i] = 0;
        path[i] = -1;
    }

    visited[s] = true;
    dist[s] = 0;
    int newp = s;
    for(int i = 0; i < n-1; i++)
    {
        //松弛操作
        for(int j = 0; j < edges[newp].size(); j++)
        {
            int next = edges[newp][j].next;
            int len = edges[newp][j].dis;
            int weight = edges[newp][j].cost;

            //最短路径不止一条
            if(dist[next] == dist[newp]+len)
            {
                if(cost[next] > cost[newp] + weight)
                {
                    cost[next] = cost[newp] + weight;
                    path[next] = newp;
                }
            }

            if(dist[next] == -1 || dist[newp]+len < dist[next])
            {
                dist[next] = dist[newp] + len;
                cost[next] = cost[newp] + weight;
                path[next] = newp;
            }
        }
        //找下一个newp
        int min = 111111111;
        for(int j = 0; j < n; j++)
        {
            if(visited[j] || dist[j] == -1)
                continue;
            if(dist[j] < min)
            {
                min = dist[j];
                newp = j;
            }
        }
        visited[newp] = true;
    }
}

int main()
{
    int n, m, s, d;
    cin >> n >> m >> s >> d;
    int a, b, x, y;
    for(int i = 0; i < m; i++)
    {
        Node tmp;
        cin >> a >> b >> x >> y;
        tmp.next = b, tmp.dis = x, tmp.cost = y;
        edges[a].push_back(tmp);
        tmp.next = a;
        edges[b].push_back(tmp);
    }
    
    Dij(n, s);
    print(d);
    cout << dist[d] << " " << cost[d] << endl;
    return 0;
}