# ACM-ICPC 2018 南京赛区网络预赛-L-Magical Girl Haze-（分层最短路）

xiaoxiao2021-03-01  20

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.

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

3

ACM-ICPC 2018 南京赛区网络预赛

ac：

#include<stdio.h> #include<string.h> #include<math.h> //#include<map> //#include<set> #include<deque> #include<queue> #include<stack> #include<bitset> #include<string> #include<fstream> #include<iostream> #include<algorithm> using namespace std; #define ll long long //#define max(a,b) (a)>(b)?(a):(b) //#define min(a,b) (a)<(b)?(a):(b) #define clean(a,b) memset(a,b,sizeof(a))// 水印 //std::ios::sync_with_stdio(false); const int MAXN=1e5+10; const int INF=0x3f3f3f3f; const ll mod=1e9+7; struct node{ int v,nxt; ll w; node(int _v=0,ll _w=0,int _nxt=0): v(_v),w(_w),nxt(_nxt){} }edge[MAXN<<2]; int head[MAXN],ecnt; ll dis[MAXN][110];//MAXN点，100层 int vis[MAXN][110]; int n,m,k; void intt() { clean(dis,INF); clean(vis,0); clean(head,-1); ecnt=0; } void add(int u,int v,ll w) { edge[ecnt]=node(v,w,head[u]); head[u]=ecnt++; } /*---上面的是板子，不用动---*/ void djstr() { dis[1][0]=0; priority_queue<pair<int,pair<int,int> > > que; que.push(make_pair(0,make_pair(1,0))); while(que.size()) { while(vis[que.top().second.first][que.top().second.second] &&que.size()>0) que.pop(); pair<int,int> p=que.top().second; vis[p.first][p.second]=1; for(int i=head[p.first];i+1;i=edge[i].nxt) { int temp=edge[i].v; //同层查找 if(vis[temp][p.second]==0 &&dis[temp][p.second]>dis[p.first][p.second]+edge[i].w) { dis[temp][p.second]=dis[p.first][p.second]+edge[i].w; que.push(make_pair(-dis[temp][p.second],make_pair(temp,p.second))); } //下层查找 if(p.second+1<=k&&vis[temp][p.second+1]==0 &&dis[temp][p.second+1]>dis[p.first][p.second]) { dis[temp][p.second+1]=dis[p.first][p.second]; que.push(make_pair(-dis[temp][p.second+1],make_pair(temp,p.second+1))); } } } } int main() { int T; scanf("%d",&T); while(T--) { intt(); scanf("%d%d%d",&n,&m,&k); int a,b; ll l; for(int i=1;i<=m;++i) { scanf("%d%d%lld",&a,&b,&l); add(a,b,l); } djstr(); printf("%lld\n",dis[n][k]); } }