L. Magical Girl Haze

There are NN cities in the country, and MM directional 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.

樣例輸入複製

1
5 6 1
1 2 2
1 3 4
2 4 3
3 4 1
3 5 6
4 5 2

樣例輸出複製

3

題目來源

ACM-ICPC 2018 南京賽區網絡預賽

 

思路：網上有兩種思路 ，大體出發點一樣 ，因爲K<=10，所以思想就是將原有圖擴充爲十層，第j層表示j次邊化爲零，跑一個spfa，自己笨啊~不會啊~難受啊~

 

代碼：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=5e6+5;

struct edge{
    ll v,w,nex;
    edge(ll v=0,ll w=0,ll nex=0):v(v),w(w),nex(nex){}
}e[maxn];

ll p[maxn];
ll en;

void add(ll u,ll v,ll w){
    e[++en]=edge(v,w,p[u]);
    p[u]=en;
}

struct node{
    ll id,w;
    node(ll id,ll w):id(id),w(w){}

    bool operator <(const node &a)const{
        return w>a.w;
    }
};

bool vis[maxn];
long long dis[maxn];

void dij(){
    priority_queue<node>q;
    memset(dis,0x3f3f3f3f,sizeof(dis));
    memset(vis,false,sizeof(vis));
    dis[1]=0;
    q.push(node(1,0));
    while(!q.empty()){
        node tmp=q.top();
        q.pop();
        long long id=tmp.id;
//        printf("%lld %lld \n",id,dis[id]);
        vis[id]=true;
        for(long long i=p[id];~i;i=e[i].nex){
            long long v=e[i].v;
            long long val=e[i].w;
            if(vis[v]) continue;
            if(dis[v]>dis[id]+val){
                dis[v]=dis[id]+val;
                q.push(node(v,dis[v]));
            }
        }
    }
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        memset(p,-1,sizeof(p));
        en=-1;
        ll u,v,w;
        for(int i=1;i<=m;i++){
            scanf("%lld%lld%lld",&u,&v,&w);
            for(int j=0;j<=k;j++){
                add(u+n*j,v+n*j,w);
                if(j!=k) add(u+n*j,v+n*(j+1),0);
            }
        }
        dij();
        printf("%lld\n",dis[n*(k+1)]);
    }
}