There are $N$ cities in the country, and $M$ directional roads from $u$ to $v(1\le u, v\le n)$. Every road has a distance $c_i$ . Haze is a Magical Girl that lives in City 1, she can choose no more than $K$ roads and make their distances become 0. Now she wants to go to City $N$, please help her calculate the minimum distance.

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

For each test case, the first line has three integers $N, M$ and $K$.

Then the following M lines each line has three integers, describe a road, $U_i, V_i, C_i$. There might be multiple edges between $u$ and $v$.

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

Output
For each test case, print the minimum distance.

样例输入
1
5 6 1
1 2 2
1 3 4
2 4 3
3 4 1
3 5 6
4 5 2
样例输出
3

思路：DP+最短路。$d[i][j]$表示到达第$i$个城市，且已经使用了$j$次机会的最短距离。

然后其它的就和最短路一样求就行了。

#include<bits/stdc++.h>
const int MAX=2e5+10;
const int MOD=1e9+7;
using namespace std;
typedef long long ll;
vector<pair<int,int> >e[MAX];
struct lenka
{
    ll now,cost,t;
    bool operator<(const lenka& p)const{return p.cost<cost;}
};
priority_queue<lenka>p;
int Time;
ll d[MAX][14];
void bfs()
{
    d[1][0]=0;
    p.push({1,0,0});
    while(!p.empty())
    {
        lenka A=p.top();p.pop();
        if(A.cost!=d[A.now][A.t])continue;
        for(int i=0;i<e[A.now].size();i++)
        {
            pair<int,int>nex=e[A.now][i];
            if(d[nex.first][A.t]==-1||d[nex.first][A.t]>A.cost+nex.second)//不把这条边的权值变为0
            {
                d[nex.first][A.t]=A.cost+nex.second;
                p.push({nex.first,A.cost+nex.second,A.t});
            }
            if(A.t+1<=Time)
            {
                if(d[nex.first][A.t+1]==-1||d[nex.first][A.t+1]>A.cost)//把这条边的权值变为0
                {
                    d[nex.first][A.t+1]=A.cost;
                    p.push({nex.first,A.cost,A.t+1});
                }
            }
        }
    }
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=n;i++)e[i].clear();
        for(int i=1;i<=m;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            e[x].push_back({y,z});
        }
        Time=k;
        for(int i=1;i<=n;i++)
        for(int j=0;j<=Time;j++)d[i][j]=-1;
        bfs();
        ll ans=1e18;
        for(int i=0;i<=Time;i++)
        {
            if(d[n][i]!=-1)ans=min(ans,d[n][i]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}