※NC14700 追债之旅

最短路变形题,加一维状态就好啦

\(dist[i][j]\)表示到达第\(i\)号点,到达时间为第\(j\)天的最短距离

  • 判重数组要和\(dist\)数组一致

直接跑\(dijkstra\)即可

const int N=1010;
vector<PII> g[N];
struct Node
{
    int dis,u,day;
    bool operator>(const Node &W) const
    {
        return dis>W.dis;
    }
};
int dist[N][15];
bool vis[N][15];
int cost[15];
int n,m,k;

void dijkstra()
{
    memset(dist,0x3f,sizeof dist);
    priority_queue<Node,vector<Node>,greater<Node> > heap;
    dist[1][0]=0;
    heap.push({0,1,0});

    while(heap.size())
    {
        int t=heap.top().u,d=heap.top().day;
        heap.pop();

        if(vis[t][d]) continue;
        vis[t][d]=true;

        for(int i=0;i<g[t].size();i++)
        {
            int j=g[t][i].fi,w=g[t][i].se;
            if(d+1<=k && dist[j][d+1] > dist[t][d] + w + cost[d+1])
            {
                dist[j][d+1]=dist[t][d]+w+cost[d+1];
                heap.push({dist[j][d+1],j,d+1});
            }
        }
    }
}

int main()
{
    cin>>n>>m>>k;

    while(m--)
    {
        int a,b,c;
        cin>>a>>b>>c;
        g[a].pb({b,c});
        g[b].pb({a,c});
    }

    for(int i=1;i<=k;i++) cin>>cost[i];

    dijkstra();

    int res=INF;
    for(int i=1;i<=k;i++)
        res=min(res,dist[n][i]);
    if(res == INF) puts("-1");
    else cout<<res<<endl;

    //system("pause");
}
posted @ 2020-10-10 11:36  Dazzling!  阅读(99)  评论(0编辑  收藏  举报