模板：

注意这个只是单向的双向的需要在更新一次

#include<bits/stdc++.h>using namespace std;typedef long long ll;
typedef pair<int, int>PII;
const int N=2e5+10;
const int MOD = 998244353;
const int INF=0X3F3F3F3F;
const int dx[]={-1,1,0,0,-1,-1,+1,+1};
const int dy[]={0,0,-1,1,-1,+1,-1,+1};
const int M = 1e6 + 10;//好累啊
//朴素版的迪杰斯特拉算法int n, m;
int g[2010][2010];
int dis[N];
bool st[N];int dij()
{memset(dis, 0x3f, sizeof dis);dis[1] = 0;for(int i = 0; i < n; i ++){int t = -1;for(int j = 1; j <= n; j ++){if(!st[j] && (t == -1 || dis[t] > dis[j])) t = j;//找到下一个最短的来进行更新}st[t]  = true;//标记一下已经走过了for(int j = 1; j <= n; j ++){dis[j] = min(dis[t] + g[t][j], dis[j]);//从那个最短端点依次更新}}if(dis[n] == 0x3f3f3f3f) return -1;return dis[n];
}
int main()
{cin >> n >> m;memset(g, 0x3f, sizeof g);while(m --){int a, b, c;scanf("%d%d%d", &a, &b, &c);g[a][b] = min(g[a][b], c);}cout << dij() << endl;return 0;
}