Graph Theory Training in Eleven

https://www.luogu.org/problem/P2939

Figure shortest path problem layered board , for a long time I did not write the shortest path, and be review about it

According to the meaning of problems, we can find the range of k is relatively small, so you can directly use layered graph

To a point force is split into k a, on behalf of the original layer using 0 chance to upgrade road, other figures indicate the use of 1, 2 ... k chance upgrade path, then the edge can be connected . Consider the relationship between each layer, the i-th layer and the right side of the first layer i + 1 is 0, spent equal opportunity to upgrade path.

The final running side dijktra, then the value of minimum distance, This question is finished it

code：

#include<bits/stdc++.h>
using namespace std;
const int maxn=100100;
const int maxm=500500;
int nextt[maxm*42],w[maxm*42],to[maxm*42],head[maxn*42],cnt=0;
void add(int u,int v,int cost)
{
    cnt++;
    nextt[cnt]=head[u];
    head[u]=cnt;
    to[cnt]=v;
    w[cnt]=cost;
}
struct node
{
    int u,dis;
    bool operator<(const node x) const
    {
        return dis>x.dis;
    }
};
priority_queue<node> q;
int dist[maxn*21];

void dij(int s)
{
    memset(dist,0x3f,sizeof(dist));
    dist[s]=0;
    q.push((node){s,0});
    while (!q.empty())
    {
        node fr=q.top();q.pop();
        int u=fr.u,dis=fr.dis;
        if (dis!=dist[u]) continue;
        for (int v=head[u];v;v=nextt[v])
            if (dist[to[v]]>dist[u]+w[v])
            {
                dist[to[v]]=dist[u]+w[v];
                q.push((node){to[v],dist[to[v]]});
            }
    }
}
int n,m,k;
int main()
{
    cin>>n>>m>>k;
    for (int i=1;i<=m;i++)
    {
        int u,v,cost;
        cin>>u>>v>>cost;
        add(u,v,cost);add(v,u,cost);
        for (int j=1;j<=k;j++)
        {
            add(n*j+u,n*j+v,cost);add(n*j+v,n*j+u,cost);
            add(n*(j-1)+u,n*j+v,0);add(n*(j-1)+v,n*j+u,0);
        }
    }
    dij(1);
    int ans=dist[n];
    for (int i=1;i<=k;i++)
        ans=min(ans,dist[i*n+n]);
    cout<<ans<<endl;
    return 0;
}