Lesson 3: SPFA seeking the shortest path

subject ：AcWing 851. spfa Find the shortest path
Given a n A little bit m Directed graph of strip edge , There may be double edges and self rings in the graph , The edge weight may be negative .

Please find out 1 It's on the n The shortest distance of point No , If you can't get from 1 Go to No n Number point , The output impossible.

The data guarantees that there is no negative weight loop .

Input format
The first line contains integers n and m.

Next m Each row contains three integers x,y,z, Indicates that there is a from point x point-to-point y The directed side of , Side length is z.

Output format
Output an integer , Express 1 It's on the n The shortest distance of point No .

If the path doesn't exist , The output impossible.

Data range
1≤n,m≤10⁵,
The absolute value of side length involved in the figure shall not exceed 10000.

sample input ：

3 3
1 2 5
2 3 -3
1 3 4

sample output ：

2

#include <iostream>
#include <cstring>
#include <queue>
#include <map>
#include <vector>

using namespace std;
typedef pair<int,int> PII;
map<int,vector<PII>> list;
const int N = 100010;
int n,m;
int dist[N],st[N];

void spfa()
{
    
    memset(dist,0x3f,sizeof(dist));
    dist[1]=0;
    queue<int> q;
    q.push(1);
    st[1]=true;
    while(q.size())
    {
    
        int t=q.front();
        q.pop();
        st[t]=false;
        for(auto it:list[t])
        {
    
            if(dist[it.first]>dist[t]+it.second)
            {
    
                dist[it.first]=dist[t]+it.second;
                if(!st[it.first])
                {
    
                    q.push(it.first);
                    st[it.first]=true;
                }
            }
        }
    }
}
int main()
{
    
    cin>>n>>m;
    while(m--)
    {
    
        int a,b,c;
        cin>>a>>b>>c;
        list[a].push_back({
    b,c});
    }
    spfa();
    if(dist[n]==0x3f3f3f3f)cout<<"impossible"<<endl;
    else cout<<dist[n]<<endl;
    return 0;
}