https://nanti.jisuanke.com/t/31001

There are NN cities in the country, and MMdirectional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distances become 00. Now she wants to go to City NN, please help her calculate the minimum distance.

Input

The first line has one integer T(1 \le T\le 5)T(1≤T≤5), then following TT cases.

For each test case, the first line has three integers N, MN,M and KK.

Then the following MM lines each line has three integers, describe a road, U_i, V_i, C_iUi,Vi,Ci. There might be multiple edges between uu and vv.

It is guaranteed that N \le 100000, M \le 200000, K \le 10N≤100000,M≤200000,K≤10,
0 \le C_i \le 1e90≤Ci≤1e9. There is at least one path between City 11 and City NN.

Output

For each test case, print the minimum distance.

这个题目跟普通最短路不同的是可以有k条边的距离缩短成0，所以需要拆点，把每个点拆成k个，数组dist[i][k]表示起点到第i个点去掉k条边后的最短路程

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 10000000000000000LL
const int maxn=100010;
int K,N,m;
struct qnode
{
    int v;
    int tt;
    ll c;
    qnode(int _v=0,int _tt=0,int _c=0):v(_v),tt(_tt),c(_c){}
    bool operator<(const qnode &r)const{
        return c>r.c;
    }
};
struct edge
{
    int v,cost;
    edge(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<edge>E[maxn];
bool vis[maxn][13];
ll dist[maxn][13];
void Dijkstra(int n,int start)//点的编号从1开始
{
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++)
        for(int k=0;k<=K;k++)
        dist[i][k]=inf;
    priority_queue<qnode>que;
    while(!que.empty())que.pop();
    dist[start][0]=0;
    que.push(qnode(start,0,0));
    qnode tmp;
    while(!que.empty())
    {
        tmp=que.top();
        que.pop();
        int k=tmp.tt;
        int u=tmp.v;
        if(vis[u][k])continue;
        vis[u][k]=true;
        for(int i=0;i<E[u].size();i++)
        {
            int v=E[tmp.v][i].v;
            int cost=E[u][i].cost;
            if(!vis[v][k]&&dist[v][k]>dist[u][k]+cost)
            {
                dist[v][k]=dist[u][k]+cost;
                que.push(qnode(v,k,dist[v][k]));
            }
            if(!vis[v][k+1]&&k<K&&dist[v][k+1]>dist[u][k])
            {
                dist[v][k+1]=dist[u][k];    //因为没有加cost,相当于cost为0，去掉一条边
                que.push(qnode(v,k+1,dist[v][k+1]));
            }
        }
    }
}
void addedge(int u,int v,int w)
{
    E[u].push_back(edge(v,w));
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&N,&m,&K);
        for(int i=1;i<=N;i++)
            E[i].clear();
        int u,v,w;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
        }
        Dijkstra(N,1);
        ll ans=inf;
        for(int i=0;i<=K;i++)
            ans=min(ans,dist[N][i]);
        printf("%lld\n",ans);
    }
}