Dijkstra+优先队列 模板

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const int maxn=1e5+5;
 5 const ll inf=1e17;
 6 struct node
 7 {
 8     ll dis;
 9     int num,pos;
10     node() {}
11     node(ll dis,int num,int pos):dis(dis),num(num),pos(pos) {}
12     bool operator< (const node& a)const
13     {
14         return dis>a.dis;
15     }
16 };
17 struct edge
18 {
19     int to,next;
20     ll z;
21 } e[maxn*2];//双边，无向图，所以乘以2
22 int head[maxn],cnt;
23 void add(int x,int y,ll w)// w边值
24 {
25     e[cnt].to=y;
26     e[cnt].z=w;
27     e[cnt].next=head[x];
28     head[x]=cnt++;
29 }
30 ll dist[maxn][12],vis[maxn][12];
31 
32 int n,m,k;
33 void dijkstra()
34 {
35     priority_queue<node>Q;//优先队列
36     Q.push(node(0,0,1));
37     while(!Q.empty())
38     {
39         node v=Q.top();
40         Q.pop();
41         if(vis[v.pos][v.num])
42             continue;
43         vis[v.pos][v.num]=1;
44         for(int i=head[v.pos]; ~i; i=e[i].next)
45         {
46             int ne=e[i].to;
47             if(dist[ne][v.num]>v.dis+e[i].z)
48             {
49                 dist[ne][v.num]=v.dis+e[i].z;
50                 Q.push(node(dist[ne][v.num],v.num,ne));
51             }
52             if(v.num<k && v.dis<dist[ne][v.num+1])
53             {
54                 dist[ne][v.num+1]=v.dis;
55                 Q.push(node(v.dis,v.num+1,ne));
56             }
57         }
58     }
59 }
60 void init()//初始化图，有点之间距离为无穷，每个点的标志初始为0，边的条数cnt初始为0
61 {
62     for(int i=0; i<maxn; i++)
63     {
64         head[i]=-1;
65         for(int j=0; j<=10; j++)
66             dist[i][j]=inf,vis[i][j]=0;
67     }
68     cnt=0;
69 }
70 int main()
71 {
72     int t;
73     scanf("%d",&t);
74     while(t--)
75     {
76         scanf("%d%d%d",&n,&m,&k);
77         int x,y;
78         ll z;
79         init();
80         for(int i=1; i<=m; i++)
81         {
82             scanf("%d%d%lld",&x,&y,&z);
83             add(x,y,z);
84         }
85         dist[1][0]=0;
86         dijkstra();
87         ll ans=inf;
88         for(int i=0; i<=k; i++)
89             ans=min(ans,dist[n][i]);
90         printf("%lld\n",ans);
91 
92     }
93     return 0;
94 }