最短路：spfa算法

• • 题目描述
  • 参考代码

    题目描述

    参考代码

    输入示例

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

    输出示例

    2
    

    #include 
    #include 
    #include 
    #include 
    using namespace std;
    const int N = 1e5 + 10;
    int n, m;
    int h[N], e[N], ne[N], w[N], idx;
    int dist[N];
    bool st[N];
    void add(int a, int b, int c)
    {
        e[idx] = b; ne[idx] = h[a]; w[idx] = c; h[a] = idx; idx++;
    }
    int spfa()
    {
        memset(dist, 0x3f, sizeof dist);
        dist[1] = 0;
        
        queue q;
        q.push(1);
        st[1] = true;
        
        while (q.size())
        {
            auto t = q.front();
            q.pop();
            
            st[t] = false;
            
            for (int i = h[t]; i != -1; i = ne[i])
            {
                int j = e[i];
                if (dist[j] > dist[t] + w[i])
                {
                    dist[j] = dist[t] + w[i];
                    if (!st[j])
                    {
                        q.push(j);
                        st[j] = true;
                    }
                }
            }
        }
        
        if (dist[n] == 0x3f3f3f3f) return -1;
        return dist[n];
    }
    int main()
    {
        scanf("%d%d", &n, &m);
        memset(h, -1, sizeof h);
        
        while (m -- )
        {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            add(x, y, z);
        }
        
        int t = spfa();
        
        if (t == -1) printf("impossible\n");
        else printf("%d\n", t);
        
        return 0;
    }