2019牛客多校第四场 J-free (Dijkstra+dp)

链接：https://ac.nowcoder.com/acm/contest/884/J

Your are given an undirect connected graph.Every edge has a cost to pass.You should choose a path from S to T and you need to pay for all the edges in your path. However, you can choose at most k edges in the graph and change their costs to zero in the beginning. Please answer the minimal total cost you need to pay.

输入描述:

The first line contains five integers n,m,S,T,K.

For each of the following m lines, there are three integers a,b,l, meaning there is an edge that costs l between a and b.

n is the number of nodes and m is the number of edges.

输出描述:

An integer meaning the minimal total cost.

示例1

输入

复制

3 2 1 3 1
1 2 1
2 3 2

输出

复制

1

备注:

1≤n,m≤103,1≤S,T,a,b≤n,0≤k≤m,1≤l≤1061≤n,m≤103,1≤S,T,a,b≤n,0≤k≤m,1≤l≤106.
Multiple edges and self loops are allowed.

 1 #include <iostream>
 2 #include <queue>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 const int N = int(1e3) + 10;
 7 const int INF = 0x3f3f3f3f;
 8 
 9 struct Edge{
10     int to,w,next;
11 }e[N+N];
12 
13 int cnt,head[N],K;
14 
15 void add(int a,int b,int w){
16     int id = ++cnt;
17     e[id].to=b;
18     e[id].w=w;
19     e[id].next=head[a];
20     head[a]=id;
21 }
22 
23 struct Node{
24     int num,dis,k;
25     bool operator < (const Node& x)const{
26         return dis > x.dis;
27     }
28 };
29 
30 int vis[N][N],dis[N][N];
31 
32 void Dijkstra(int s){
33     priority_queue<Node> q;
34     memset(vis, 0, sizeof(vis));
35     memset(dis, INF, sizeof(dis));
36     dis[s][0]=0;
37     q.push(Node{s,dis[s][0],0});
38     while(!q.empty()){
39         Node x =q.top();
40         q.pop();
41         int n=x.num;
42         int k=x.k;
43         if(vis[n][k]) continue;
44         vis[n][k]=1;
45         for(int i =head[n]; i != -1; i=e[i].next){
46             int v=e[i].to,w=e[i].w;
47             if(k < K && dis[n][k] < dis[v][k+1]){
48                 dis[v][k+1]=dis[n][k];
49                 q.push(Node{v,dis[v][k+1],k+1});
50             }
51             if(dis[n][k] + w < dis[v][k]){
52                 dis[v][k] = dis[n][k] + w;
53                 q.push(Node{v,dis[v][k],k});
54             }
55         }
56     }
57 }
58 
59 int main(){
60     ios::sync_with_stdio(0);
61     memset(head,-1,sizeof(head));
62     int n,m,S,T;
63     cin>>n>>m>>S>>T>>K;
64     while(m--){
65         int a,b,l;
66         cin>>a>>b>>l;
67         add(a,b,l);
68         add(b,a,l);
69     }
70     Dijkstra(S);
71     int ans=INF;
72     for(int i=0;i<=K;i++) ans=min(ans,dis[T][i]);
73     cout<<ans<<endl;
74 }