步骤

1. 初始化dist[], dist[1] = 0, 其余设置为∞
2. n次迭代dist
3. t=-1时，t=j=1, 更新n个点的dist[]
4. 求出比t=1距离更小的点，即离起点最近的点
5. 用4中的点去更新dist[]

#include <iostream>
#include <cstring>

using namespace std;

const int N = 510;

int dist[N];
int st[N];
int n, m;
int g[N][N];

int dijkstra()
{
    // 初始化dist[]
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    for (int i = 0; i < n; i ++)
    {
        int t = -1; // A
        for (int j = 1; j <= n; j ++)
        {
            if (!st[j] && (t == -1 || dist[j] < dist[t]))
                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;
    return dist[n];
}

int main()
{
    // int a, b, c;
    cin >> n >> m;
    memset(g, 0x3f, sizeof g);
    for (int i = 1; i <= m; i ++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        // add(a, b, c);
        g[a][b] = min(g[a][b], c);
    }
    cout << dijkstra() << endl;
    return 0;
}