PAT A1030 Dijsktra I'm sorry I could not even write your name right

Ow Ow did not expect to finish vowed independent Dijsktra algorithm or write to
I can not even have your name misspelled, pop ......
This question needs D algorithm and DFS algorithm combines ......
Yes I have forgotten the DFS hehe

#include<iostream>
#include<string>
#include<queue>
using namespace std;

#define INF 99999999
int road[500][500];
int cost[500][500] = { 0 };
int w[500] = { 0 };
int vis[500] = { 0 };
int dis[500];
int pre[500];
int N, M, start, des;//城市数目，高速数目，起点，终点

void Dijkstra(int st, int num)
{
	dis[st] = 0;
	for (int i = 0; i < num; i++)
	{
		int u = -1, MIN = INF;
		for (int j = 0; j < num; j++)
		{
			if (vis[j] == 0 && dis[j] < MIN)
			{
				u = j;
				MIN = dis[j];
			}
		}
		if (u == -1) return;
		vis[u] = 1;
		for (int v = 0; v < num; v++)
		{
			if (vis[v] == 0 && road[u][v] != INF)
			{
				if (dis[u] + road[u][v] < dis[v])
				{
					dis[v] = dis[u] + road[u][v];
					w[v] = w[u] + cost[u][v];
					pre[v] = u;
				}
				else if (dis[u] + road[u][v] == dis[v] && w[u] + cost[u][v] < w[v])
				{
					w[v] = w[u] + cost[u][v];
					pre[v] = u;
				}
			}
		}
	}
}

void Dfs(int fina)
{
	if (fina == start)
	{
		return;
	}
	Dfs(pre[fina]);
	cout << pre[fina] << " ";
}
int main()
{

	cin >> N >> M >> start >> des;
	fill(road[0], road[0] + 500 * 500, INF);
	fill(dis, dis + 500, INF);
	int s, d;
	while (M--)
	{
		cin >> s >> d;
		cin >> road[s][d] >> cost[s][d];
		road[d][s] = road[s][d];
		cost[d][s] = cost[s][d];
	}
	Dijkstra(start, N);
	Dfs(des);
	cout << des<<" "<<dis[des] << " " << w[des];
	return 0;

}

Then gradually I realized ...... scanf_s printf_s and easy to use, too incense ......
to a D algorithm template ......

for(从头到尾）
{
	定义两个U变量 u=-1;MIN=INF;
	for(从头到尾）
	{
		if(vis==0&&dis<MIN)
			u=...
	}
	u的判定及定义
	u=-1就return 反正 vis[u]=1
	for(从头到尾）
	{
		筛选条件
    }
}

Ha ha ha ha ha I do not believe I can not remember ......