源点-起点 汇点-终点

迪杰斯特拉朴素版
1.初始化dist起点为0 其余点为无穷大
2.重复n次 每次找到未确定的离起点最短的点t 用t对其它点的距离进行更新
3.将点t标记为已确定

[迪杰斯特拉朴素版] [c++]
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#include<bits/stdc++.h>
using namespace std;

const int N=520;
int n,m,dist[N],g[N][N];
bool st[N];

int dj()
{
    memset(dist,0x3f,sizeof dist); 
    dist[1]=0; //除了起点都初始化为正无穷
    for(int i=0;i<n;i++)
    {
        int t=-1;
        for(int j=1;j<=n;j++)
        {
            if(!st[j] &&(t==-1 || dist[t]>dist[j])) //找t 即dist最小值 
            {
                t=j; 
            }
        }
        st[t]=true;
        for(int j=1;j<=n;j++)
        {
            dist[j]=min(dist[j],dist[t]+g[t][j]);
        }
    }
    if(dist[n]==0x3f3f3f3f) return -1;
    else return dist[n];
}

int main()
{
    cin>>n>>m;
    memset(g,0x3f,sizeof g);
    while(m--)
    {
        int a,b,c;
        cin>>a>>b>>c;
        g[a][b]=min(g[a][b],c);
    }
    int t=dj();
    cout<<t<<endl;
    return 0;
}


很简单 n次迭代 循环所有边 如果第n次仍松弛 说明有负权回路

[bellman-ford] [c++]
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#include <bits/stdc++.h>
using namespace std;
const int N = 510, M = 1e4 + 10;
int n, m, k;
int dist[N], back[N];
struct Edge {
    int a, b, c;
} edge[M];

int bellman_ford() {
    dist[1] = 0;
    for (int i = 1; i <= k; i++) {  //此处k 经过不超过k条边的最短路距离
        memcpy(back, dist, sizeof dist);
        for (int j = 1; j <= m; j++) {
            int a = edge[j].a, b = edge[j].b, w = edge[j].c;
            dist[b] = min(dist[b], back[a] + w);
        }
    }
    if (dist[n] >= 0x3f3f3f3f / 2) return -0x3f3f3f3f;
    return dist[n];
}

int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 1; i <= m; i++) {
        int u, v, w; scanf("%d%d%d", &u, &v, &w);
        edge[i] = {u, v, w};
    }
    memset(dist, 0x3f3f3f3f, sizeof dist);
    int ans = bellman_ford();
    if (ans == -0x3f3f3f3f) puts("impossible");
    else printf("%d", ans);
    return 0;
}