Dijkstra single source shortest circuit template

Let's update the concave corner of the radar map.\(Dijkstra\)The shortest circuit of a single source can be handled, that is, the shortest distance between a specified point and each point can be obtained once. Negative edge rights cannot be handled.

#include <bits/stdc++.h>
 using namespace std;

 int pre[100010], k; //Save pictures
 bool vis[100010];//Have you visited it
 int n, m, s; //Input

 struct edge {
	 int to, nxt, len; //The chained front list star
 } e[1010100];

 struct node {
	 int pos, dis; //number, distance
	 bool operator < (const node &cmp) const {
		 return dis > ;//Reload small root heap
	 }
 } dis[1010100];
 priority_queue <node> q;

 void add(int u, int v, int l) {
	 e[++k] = {v, pre[u], l};
	 pre[u] = k;
 }

 void dijkstra() {
	 dis[s].dis = 0;//Don't forget
	 ({s, 0}); //Enter the team from the starting point
	 while (!()) {
		 node tmp = ();//Prefer to the point with the smallest distance
		 ();//Don't forget this or it will pass away. You have to pop the number from the beginning, otherwise it is possible to pop the number you just pushed.
		 If (vis[]) continue; //Put off visited points first
		 vis[] = 1;
		 //Start relaxation of this node
		 for (int i = pre[]; i; i = e[i].nxt) {
			 //Transfer all neighboring nodes of this node
			 int to = e[i].to;
			 if (dis[to].dis > dis[].dis + e[i].len) {
				 //If the distance from s to the current traversal node is greater than the distance from tmp node to the current traversal node, the distance can be updated
				 dis[to].dis = dis[].dis + e[i].len;
				 ({to,dis[to].dis});
			 }
		 }
	 }
 }

 int main() {
	 scanf("%d%d%d", &n, &m, &s);
	 for (int i = 1; i <= n; i++) dis[i].dis = 2147483647;//Cancer, if there is no solution, you need to output 2^31-1
	 for (int i = 1; i <= m; i++) {
		 int u, v, l;
		 scanf("%d%d%d", &u, &v, &l);
		 add(u, v, l); //Directed
	 }
	 dijkstra();
	 for (int i = 1; i <= n; i++) {
		 printf("%d", dis[i].dis);
	 }
 }