ACM-ICPC 2018 南京赛区网络预赛 L. Magical Girl Haze

Posted 2021-01-04 tingtin

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There are NN cities in the country, and MM directional roads from uu to v(1le u, vle 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 Tle 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.

样例输入复制

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

样例输出复制

3

题目来源

ACM-ICPC 2018 南京赛区网络预赛

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <utility>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <queue>
 9 #include <stack>
10 using namespace std;
11 #define  N  100009
12 #define  M  200009
13 #define lowbit(x) x&(-x)
14 #define ll long long
15 const  ll inf =9e18; 
16 int head[N],t,n,m,k,cnt;
17 ll dp[N][15];
18 struct Edge{
19 int from,to,nex;
20 ll w;    
21 }e[M*2];//当然本题不用*2(有向图)
22 struct Node{
23     int a,b;// a:终点  ，b : 已经用了几次免费路
24     ll dis;//1到a的当前最短距离
25     bool operator <(const Node &p)const{
26         return  dis>p.dis;//因此下面只能用priority_queue<Node>que;
27         //return  dis<p.dis; 是错的，不可以这么定义
28     }
29 };
30 void  init()
31 {
32     for(int i=0;i<=n;i++){
33         head[i]=-1;
34     }
35     cnt=0;
36 }
37 void add(int u,int v,ll val)// ll val
38 {
39     e[cnt].from=u;
40     e[cnt].to=v;
41     e[cnt].nex=head[u];
42     e[cnt].w=val;
43     head[u]=cnt++;
44 }
45 void bfs()
46 {
47     for(int i=0;i<=n;i++)
48     {
49         for(int j=0;j<=15;j++){
50             dp[i][j]=inf;
51         }
52     }
53     dp[1][0]=0;//dp[i][j] :从1到i ，已经有j次免费的路的最短路径
54     priority_queue<Node>que;
55     que.push(Node{1,0,0});
56     while(!que.empty()){
57         Node tmp=que.top();
58         que.pop();
59         int u=tmp.a;
60         int b=tmp.b;
61         for(int i=head[u];i!=-1;i=e[i].nex){
62             int v=e[i].to;
63         if(dp[v][b]>tmp.dis+e[i].w){//这条路不当作免费路
64             dp[v][b]=tmp.dis+e[i].w;
65             que.push(Node{v,b,dp[v][b]});            
66         }
67         if(b+1<=k){
68             if(dp[v][b+1]>tmp.dis){//这条路当作免费路
69                 dp[v][b+1]=tmp.dis;
70                 que.push(Node{v,b+1,tmp.dis});
71             }
72         }
73     }
74 }
75 }
76 int   main()
77 {
78     scanf("%d",&t);
79     while(t--)
80     {
81         scanf("%d%d%d",&n,&m,&k);
82         init();
83         int u,v;
84         ll w;
85         for(int  i=0;i<m;i++)
86         {
87             scanf("%d%d%lld",&u,&v,&w);
88             add(u,v,w);
89         }
90         bfs();
91         printf("%lld
",dp[n][k]);
92     }
93     return  0;
94 }