1 条题解

  • 0
    @ 2024-7-25 13:41:04

    参考答案:

    #include<iostream>
    #include<cstring>
    
    using namespace std;
    
    const int N = 30010;
    int n, m, k;
    int dis[N], last[N];
    
    struct Edge
    {
        int a, b, c;
    }edge[N];
    
    int bellman_ford()
    {
        memset(dis, 0x3f, sizeof(dis));
        dis[1] = 0;
    
        for (int i = 1; i <= k; ++i)
        {
            memcpy(last, dis, sizeof(dis));
            for (int j = 1; j <= m; ++j)
            {
                auto t = edge[j];
                dis[t.b] = min(dis[t.b], last[t.a] + t.c);
            }
        }
    
        return dis[n];
    }
    int main()
    {
        cin >> n >> m >> k;
    
        for (int i = 1; i <= m; ++i)
        {
            cin >> edge[i].a >> edge[i].b >> edge[i].c;
        }
    
        if (bellman_ford() > 0x3f3f3f3f / 2) cout << "No path";
        else cout << dis[n];
    
        return 0;
    }
    
    • 1

    信息

    ID
    97
    时间
    1000ms
    内存
    256MiB
    难度
    10
    标签
    递交数
    4
    已通过
    1
    上传者