题解 | #【模板】单源最短路2#

要考虑到案例2的情况。循环要到N

#include<iostream>
#include<cstring>
using namespace std;

const int N = 5100;
int n, m;
int g[N][N];
int dist[N];
bool st[N];

int dijkstra(){
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    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]))
                t = j;
        }
        st[t] = true;
        for(int j = 1; j <= N; j ++){
            dist[j] = min(dist[j], g[t][j] + dist[t]);
        }
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}

int main(){
    scanf("%d%d", &n, &m);
    memset(g, 0x3f, sizeof g);
    while(m --){
        int a, b, c;
        cin >> a >> b >> c;
        g[a][b] = g[b][a] = c;
    }
    int t = dijkstra();
    printf("%d\n",t);
    return 0;
}