1030 Travel Plan Dijkstra+dfs

和1018思路如出一辙，先求最短路径，再dfs遍历

#include <iostream>
#include <cstdio>
#include <vector>
#include<algorithm>
using namespace std;
int e[][], c[][];// distence cost
vector<int> pre[];//到达当前节点的前一个节点
vector<vector<int> > st; //最短距离的路径
vector<int> path, tmppath;
int inf = ;
int mincost = inf;
int totalcost = ;
int S;//由于dfs 要用到这个变量，所以从main里提前  出发城市
void dfs(int node, int front)
{
    tmppath.push_back(node);
    int xx;
    if (front == -)
    {
        xx = ;
    }
    else
    {
        xx = c[node][front];
    }
    totalcost += xx;
    if (node == S)
    {
        if (totalcost < mincost)
        {
            mincost = totalcost;
            path.clear();
            path = tmppath;
        }
        totalcost = totalcost - xx;
    tmppath.pop_back();
        return;
    }
    for (int i = ; i < pre[node].size(); i++)
    {
        dfs(pre[node][i], node);
    }
    totalcost = totalcost - xx;
    tmppath.pop_back();
}

int main()
{
    int N, M, D;
    int a, b, tmp1, tmp2;
    fill(e[], e[] +  * , inf);
    fill(c[], c[] +  * , inf);
    scanf("%d%d%d%d", &N, &M, &S, &D);
    for (int i = ; i < M; i++)
    {
        scanf("%d%d", &a, &b);
        scanf("%d", &tmp1);
        e[a][b] = e[b][a] = tmp1;
        scanf("%d", &tmp2);
        c[a][b] = c[b][a] = tmp2;
    }       //输入结束
    //Dijkstra
    int dis[];
    int mark[];
    fill(mark, mark + , );
    fill(dis, dis + , inf);
    dis[S] = ;
    pre[S].push_back(S);
    int point, near;//当前选择要加入的点 和 最近
    for (int i = ; i < N; i++)
    {
        point = -, near = inf;
        for (int j = ; j < N; j++)
        {
            if (mark[j] ==  && dis[j] < near)
            {
                point = j;
                near = dis[j];
            }
        }
        if (near == inf)break;//所有可达点都已加入
        mark[point] = ;
        for (int j = ; j < N; j++)
        {
            if (dis[j] > dis[point] + e[point][j])
            {
                dis[j] = dis[point] + e[point][j];
                pre[j].clear();
                pre[j].push_back(point);
            }
            else if (dis[j] == dis[point] + e[point][j])
            {
                pre[j].push_back(point);
            }
        }

    }
    dfs(D, -);//pre存的是到当前节点的前一个节点，所以要倒回去
    if(path.size()>)
    for (int i = path.size() - ; i >= ; i--)
    {
        printf("%d ", path[i]);
    }
    else
    printf("%d",S);
    printf("%d %d", dis[D], mincost);
    return ;

}